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More questions on this topic: IGCSE Shape & Space: Basic Trigonometry (144) Level 2 FM Shape & Space: Basic Trigonometry (10)
26 \(ABC\) is a right-angled triangle.
Calculate the value of \(x\).
Give your answer correct to 1 decimal place. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 60.3 | M1 | for a correct trig statement, eg \(\tan x = \dfrac{7}{4}\) |
| A1 | 60.2 to 60.42 |
Other methods are possible but only award this mark at the point of an equation with \(x\) as the only unknown
7 \(ABC\) is a right-angled triangle.
Calculate the value of \(x\).
Give your answer correct to 1 decimal place. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 60.3 | M1 | for a correct trig statement, eg \(\tan x = \frac{7}{4}\) |
| A1 | 60.2 to 60.42 |
Other methods are possible but only award this mark at the point of an equation with \(x\) as the only unknown
24 \(ABC\) is a right-angled triangle.
Calculate the length of \(AB\).
Give your answer correct to 1 decimal place. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 14.3 | M1 | for a correct statement for \(AB\) using trigonometry, eg \(\tan 62 = \dfrac{AB}{7.6}\) or \((AB =)\ 7.6 \times \tan 62\) |
| A1 | answer in the range 14.28 to 14.3 |
If an answer is given in the range in working and then rounded incorrectly award full marks
8 \(ABC\) is a right-angled triangle.
Calculate the length of \(AB\).
Give your answer correct to 1 decimal place. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 14.3 | M1 | for a correct statement for \(AB\) using trigonometry, eg \(\tan 62 = \dfrac{AB}{7.6}\) or \((AB =)\ 7.6 \times \tan 62\) |
| A1 | answer in the range 14.28 to 14.3 |
If an answer is given in the range in working and then rounded incorrectly award full marks
21 The diagram shows a circle, radius \(r\) cm and two regular hexagons.
Each side of the larger hexagon \(ABCDEF\) is a tangent to the circle.
Each side of the smaller hexagon \(PQRSTU\) is a chord of the circle.
By considering perimeters, show that
\[3 \lt \pi \lt 2\sqrt{3}\](4)
| Answer | Mark | Mark scheme |
|---|---|---|
| Result shown | P1 | for process to find the length of half a side or a side or the perimeter of the smaller hexagon \(PQRSTU\), eg \(r\sin 30\ \left(= \dfrac{r}{2}\right)\) oe or \(2r\sin 30\ (= r)\) oe or \(6 \times 2r\sin 30\ (= 6r)\) oe |
| P1 | for process to find the length of half a side or a side or the perimeter of the larger hexagon \(ABCDEF\) eg Length of half side \(= r\tan 30\) or \(\dfrac{r}{\tan 60}\ \left(= \dfrac{\sqrt{3}r}{3}\right)\) oe or Length of side \(= 2r\tan 30\) or \(\dfrac{2r}{\tan 60}\) or \(\dfrac{r}{\sin 60}\) or \(\dfrac{r}{\cos 30}\ \left(= \dfrac{2\sqrt{3}r}{3}\right)\) oe or Length of perimeter \(= 6 \times 2r\tan 30\) or \(6 \times \dfrac{2r}{\tan 60}\) oe | |
| P1 | (dep P2) for process of forming a correct inequality, eg using half lengths eg \(\dfrac{r}{2} \lt \dfrac{2\pi r}{12} \lt r\tan 30\) oe or using lengths eg \(r \lt \dfrac{2\pi r}{6} \lt 2r\tan 30\) oe or using perimeters eg \(6 \times r \lt 2\pi r \lt 6 \times 2r\tan 30\) oe | |
| C1 | (dep P2) correct algebra leading to given result, \(3 \lt \pi \lt 2\sqrt{3}\) |
May use Sine Rule
or cos60 instead of sin30
May use Sine Rule
Note this mark is not for just the sight of \(\dfrac{\sqrt{3}r}{3}\) or \(\dfrac{2\sqrt{3}r}{3}\) or \(\dfrac{12\sqrt{3}r}{3}\) oe, they need to be associated with the correct length
Perimeter \(= 4\sqrt{3}r\) alone does not get this mark
10 The diagram shows two right-angled triangles.
All lengths are measured in centimetres.
Given that
\(\sin a = \tan b\)
work out the value of \(x\). (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.25 | B1 | for \((\sin a =)\ \dfrac{7 - 2x}{8}\) or \((\tan b =)\ \dfrac{1 + x}{4}\) |
| P1 | for start of a process to solve \(\dfrac{7 - 2x}{8} = \dfrac{1 + x}{4}\), eg \(4(7 - 2x) = 8(1 + x)\) or \(28 - 8x = 8 + 8x\) or \(7 - 2x = 2(1 + x)\) or \(7 - 2x = 2 + 2x\) or \(\dfrac{7 - 2x}{8} = \dfrac{2(1 + x)}{8}\) or \(\dfrac{7 - 2x}{8} = \dfrac{2x + 2}{8}\) | |
| A1 | for 1.25 oe |
Must come from a correct equation
29 \(ABC\) is a right-angled triangle.
Calculate the length of \(AB\).
Give your answer correct to 3 significant figures. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 32.2 | M1 | for a correct trig statement, eg \(28 \times \tan 49\) or \(\tan 49 = AB \div 28\) |
| A1 | Answer in the range 32.2 to 32.22 |
Can use a combination of skills but must have only one unknown in \(x\) to score this mark
If an answer is given in the range in working and then rounded incorrectly award full marks.
10 \(ABD\) is a triangle.
\(C\) is a point on \(BD\).
Work out the length of \(DC\).
Give your answer correct to 1 decimal place. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 3.1 | P1 | for using sin to find length of \(AC\), eg \(6.8 \times \sin 41\) or a full process to find \(AC\) or \((AC =)\ 4.46\ldots\) or a full process to find \(AD\) or \((AD =)\ 5.44\ldots\) |
| P1 | for a complete process to find a relationship involving \(DC\) eg \(\tan 55 = \dfrac{\text{``}4.46\ldots\text{''}}{DC}\) or \((DC =)\ \dfrac{\text{``}4.46\ldots\text{''}}{\tan 55}\) or \(\cos 55 = \dfrac{DC}{\text{``}5.44\ldots\text{''}}\) | |
| A1 | answer in the range 3.1 to 3.2 |
Accept rounded or truncated figures.
If a correct answer within the range is shown in working but is incorrectly rounded, award full marks.
22
Work out the value of \(x\).
Give your answer correct to 3 significant figures. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 8.73 | M1 | for a correct trig statement, eg \(14.5 \times \cos 53\) or \(\cos 53 = x \div 14.5\) |
| A1 | answer in the range 8.726 to 8.73 |
Can use a combination of skills but must have only one unknown in \(x\) to score this mark
If an answer is given in the range in working and then rounded incorrectly award full marks.
5
Work out the value of \(x\).
Give your answer correct to 3 significant figures. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 8.73 | M1 | for a correct trig statement, eg \(14.5 \times \cos 53\) or \(\cos 53 = x \div 14.5\) |
| A1 | answer in the range 8.726 to 8.73 |
Can use a combination of skills but must have only one unknown in \(x\) to score this mark
If an answer is given in the range in working and then rounded incorrectly award full marks.
12 \(ABC\) and \(ACD\) are right-angled triangles.
\(DC = 8\) cm
Angle \(ADC = 45^\circ\)
Angle \(ABC = 20^\circ\)
Work out the length of \(AB\).
Give your answer correct to 3 significant figures. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 23.4 | M1 | for stating that \(AC = 8\) or for a relationship that may be used to find \(AC\) eg (\(AC =\)) \(8 \times \tan 45\) or \(\tan 45 = \dfrac{AC}{8}\) |
| M1 | for relationship that may be used to find \(AB\), eg \(\sin(20) = \text{``}8\text{''} \div AB\) or (\(AB =\)) \(\dfrac{\text{``}8\text{''}}{\sin 20}\) | |
| A1 | for answer in the range 23.3 to 23.4 | |
| Alternative | ||
| M1 | for a relationship that may be used to find \(AD\) eg \(\cos(45) = 8 \div AD\) oe or (\(AD =\)) 11.3(13...) | |
| M1 | for a relationship that may be used to find \(AB\), eg \(\dfrac{AB}{\sin 45} = \dfrac{\text{``}11.3\text{''}}{\sin 20}\) | |
| A1 | for answer in the range 23.3 to 23.4 |
May be seen on diagram
May use the sine rule
If an answer is given in the range in working and then rounded incorrectly award full marks.
26 \(ABC\) is a right-angled triangle.
\(PQR\) is a right-angled triangle.
| Answer | Mark | Mark scheme |
|---|---|---|
| 17.8 | M1 | for \(\tan 56 = \dfrac{x}{12}\) or (BC =) \(12 \times \tan 56\) oe or alternative method to find BC |
| A1 | for an answer in the range 17.7 to 17.8 |
For any alternative method candidates must arrive at an equation with BC as the only unknown
If an answer in the range 17.7 to 17.8 is given in the working space then incorrectly rounded, award full marks.
| Answer | Mark | Mark scheme |
|---|---|---|
| 33.6 | M1 | for \(\cos x = \dfrac{15}{18}\) or \(\cos x = 0.83..\) or \(x = \cos^{-1} \dfrac{15}{18}\) or alternative method to find \(x\) |
| A1 | for an answer in the range 33.5 to 33.91 |
For any alternative method candidates must arrive at an equation with \(x\) as the only unknown
If an answer in the range 33.5 to 33.91 is given in the working space then incorrectly rounded, award full marks.
6 \(ABC\) is a right-angled triangle.
\(PQR\) is a right-angled triangle.
| Answer | Mark | Mark scheme |
|---|---|---|
| 17.8 | M1 | for \(\tan 56 = \dfrac{x}{12}\) or \((BC) = 12 \times \tan 56\) oe or alternative method to find \(BC\) |
| A1 | for an answer in the range 17.7 to 17.8 |
For any alternative method candidates must arrive at an equation with BC as the only unknown
If an answer in the range 17.7 to 17.8 is given in the working space then incorrectly rounded, award full marks.
| Answer | Mark | Mark scheme |
|---|---|---|
| 33.6 | M1 | for \(\cos x = \dfrac{15}{18}\) or \(\cos x = 0.83..\) or \(x = \cos^{-1} \dfrac{15}{18}\) or alternative method to find \(x\) |
| A1 | for an answer in the range 33.5 to 33.91 |
For any alternative method candidates must arrive at an equation with \(x\) as the only unknown
If an answer in the range 33.5 to 33.91 is given in the working space then incorrectly rounded, award full marks.
12 The diagram shows triangle \(ABC\).
\(ADC\) and \(DEB\) are straight lines.
\(AD = 4.4\) cm
\(BC = 8.6\) cm
\(E\) is the midpoint of \(DB\).
Angle \(CDB = 90^\circ\)
Angle \(DCB = 40^\circ\)
Work out the size of angle \(EAD\).
Give your answer correct to 1 decimal place.
You must show all your working. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 32.1 | P1 | starts process, eg \(\sin 40 = \dfrac{DB}{8.6}\) oe or for \(8.6 \times \sin 40\ (= 5.52797\ldots)\) |
| P1 | complete process to find \(ED\), eg \((8.6 \times \sin 40) \div 2\ (= 2.76\ldots)\) | |
| P1 | process to find angle \(EAD\), eg \(\tan^{-1}\left(\dfrac{\text{``}2.76\ldots\text{''}}{4.4}\right)\) or \(\tan^{-1}(\text{``}0.628\ldots\text{''})\) | |
| A1 | answer in range 32.09 to 32.2 |
Accept values rounded or truncated to 2 dp.
If an answer in the range is seen in working and then incorrectly rounded award full marks
24 \(ABC\) is a right-angled triangle.
Calculate the length of \(AB\).
Give your answer correct to 2 decimal places. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 9.85 | M1 | for \(\sin(38) = \dfrac{AB}{16}\) oe or alternative method to find \(AB\) |
| A1 | for an answer in the range 9.76 to 9.92 |
22 The diagram shows a circle, centre \(O\).
\(AB\) is the tangent to the circle at the point \(A\).
Angle \(OBA = 30^\circ\)
Point \(B\) has coordinates \((16, 0)\)
Point \(P\) has coordinates \((3p, p)\)
Find the value of \(p\).
Give your answer correct to 1 decimal place.
You must show all your working. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 2.5 | P1 | use of \(\sin 30 = \tfrac{1}{2}\) to find \(OA\ (= 8)\) or \(OAB = 90^\circ\) eg \(OA = 16\sin 30^\circ\) or right angle marked on diagram |
| P1 | recognition that equation of circle is \(x^2 + y^2 = r^2\) | |
| P1 | Correct substitution of \(p\), \(3p\) and \(r\) in \(x^2 + y^2 = r^2\) eg \(9p^2 + p^2 = OA^2\) or \((3p)^2 + p^2 = \text{``}8^2\text{''}\) | |
| A1 | for answer in the range 2.5 to 2.53 |
Accept \(3p^2 + p^2 = r^2\) for the award of this mark
Do not accept \(3p^2 + p^2 = 8^2\) for the award of this mark
Accept \(\sqrt{6.4}\) or \(\dfrac{4\sqrt{10}}{5}\)
If an answer within the given range is seen in working and rounded incorrectly award full marks.
Award 0 marks for the answer without supportive working.
5 \(ABC\) is a right-angled triangle.
Calculate the length of \(AB\).
Give your answer correct to 2 decimal places. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 9.85 | M1 | for \(\sin(38) = \dfrac{AB}{16}\) oe or alternative method to find \(AB\) |
| A1 | for an answer in the range 9.76 to 9.92 |
25 \(ABC\) is a right-angled triangle.
\(AC = 14\) cm.
Angle \(C = 90^\circ\)
size of angle \(B\) : size of angle \(A\) = 3 : 2
Work out the length of \(AB\).
Give your answer correct to 3 significant figures. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 17.3 | P1 | for full process to find either angle eg \((180 - 90) \div (2 + 3) \times 2\) or for 36 or 54 seen as an angle |
| P1 | for a correct equation using trigonometry eg \(\cos [A] = 14 \div AB\) | |
| P1 | (dep previous P mark) for rearranging their trigonometry equation to make \(AB\) the subject eg (\(AB =\)) \(\text{``}14 \div \cos 36\text{''}\) | |
| A1 | for an answer in the range 17.3 to 17.4 |
May be seen on diagram
Condone correct values if incorrectly placed.
This must be shown as an equation with all four elements (eg cos, [\(A\)], 14, \(AB\)) present.
[\(A\)] could be 36 or any angle clearly and unambiguously identified as \(A\).
This also applies to [\(B\)] with Sine.
If an answer is shown in the range in working and then incorrectly rounded award full marks.
8
Here is a right-angled triangle.
\(\cos 60^\circ = 0.5\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 1 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 8 | M1 | starts process, eg \(\cos(60) = \dfrac{4}{x}\) or \(0.5 = \dfrac{4}{x}\) oe or \(\sin 30 = \dfrac{4}{x}\) or \(\dfrac{\sin 30}{4} = \dfrac{\sin 90}{x}\) oe |
| A1 | cao |
All three elements of cos, 4, \(x\) must be present in an equation. eg \(\cos = 4/x\) is acceptable but \(\cos(4/x)\) is insufficient
6 \(ABC\) is a right-angled triangle.
\(AC = 14\) cm.
Angle \(C = 90^\circ\)
size of angle \(B\) : size of angle \(A\) = 3 : 2
Work out the length of \(AB\).
Give your answer correct to 3 significant figures. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 17.3 | P1 | for full process to find either angle eg \((180 - 90) \div (2 + 3) \times 2\) or for 36 or 54 seen as an angle |
| P1 | for a correct equation using trigonometry eg \(\cos [A] = 14 \div AB\) | |
| P1 | (dep previous P mark) for rearranging their trigonometry equation to make \(AB\) the subject eg (\(AB =\)) \(\text{``}14 \div \cos 36\text{''}\) | |
| A1 | for an answer in the range 17.3 to 17.4 |
May be seen on diagram
Condone correct values if incorrectly placed.
This must be shown as an equation with all four elements (eg cos, [\(A\)], 14, \(AB\)) present.
[\(A\)] could be 36 or any angle clearly and unambiguously identified as \(A\).
This also applies to [\(B\)] with Sine.
If an answer is shown in the range in working and then incorrectly rounded award full marks.
23 \(ABC\) is a right-angled triangle.
The length of the side \(AB\) is reduced by 1 cm.
The length of the side \(BC\) is still 7 cm.
Angle \(ACB\) is still \(90^\circ\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 50.5 | M1 | for \(\cos ABC = \dfrac{7}{11}\) (0.63…) oe |
| A1 | for answer in the range 50.4 to 50.51 |
Must be a complete statement for cos, sin or tan with all three elements present.
If an answer is in the range 50.4 to 50.51 is given in the working space then incorrectly rounded, award full marks.
| Answer | Mark | Mark scheme |
|---|---|---|
| Increase (supported) | C1 | States increase with supporting reason eg “\(\dfrac{7}{10}\) is greater than \(\dfrac{7}{11}\)” “0.636 is less than 0.7” ….“cos increases as angle decreases” “decreasing the denominator increases the value of the fraction” “angle is now 45.6” (accept 45.5 – 45.6) |
If figures are given they must be correct (truncated or rounded).
18 \(ABCDEFGH\) is a cuboid.
\(AB = 7.3\) cm
\(CH = 8.1\) cm
Angle \(BCA = 48^\circ\)
Find the size of the angle between \(AH\) and the plane \(ABCD\).
Give your answer correct to 1 decimal place. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 39.5 | P1 | for a start to a process eg, for a correct trigonometric statement, eg \(\sin 48 = \dfrac{7.3}{AC}\) or \(\cos 42 = \dfrac{7.3}{AC}\) or \(\dfrac{AC}{\sin 90} = \dfrac{7.3}{\sin 48}\) OR angle \(CAH\) unambiguously identified on a diagram |
| P1 | for a complete correct process to find \(AC\), eg (\(AC =\)) \(\dfrac{7.3}{\sin(48)}\) (=9.8..) or (\(AC =\)) \(\dfrac{7.3}{\cos(42)}\) (=9.8..) or (\(AC =\)) \(\sin 90 \times \dfrac{7.3}{\sin 48}\) (=9.8..) | |
| P1 | for a correct statement using angle \(CAH\), eg \(\tan(CAH) = \dfrac{8.1}{\text{``}9.8...\text{''}}\) OR \(\sqrt{8.1^2 + \text{``}9.8\text{''}^2}\) (=12.7…) and \(\dfrac{\sin CAH}{8.1} = \dfrac{\sin 90}{\text{``}12.7\text{''}}\) | |
| A1 | for answer in the range 39.5 – 39.51 |
P1 (start): Must include correct values
If an answer is given in the range but then incorrectly rounded award full marks.
5 \(ABC\) is a right-angled triangle.
The length of the side \(AB\) is reduced by 1 cm.
The length of the side \(BC\) is still 7 cm.
Angle \(ACB\) is still \(90^\circ\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 50.5 | M1 | for \(\cos ABC = \dfrac{7}{11}\) (0.63…) oe |
| A1 | for answer in the range 50.4 to 50.51 |
Must be a complete statement for cos, sin or tan with all three elements present.
If an answer is in the range 50.4 to 50.51 is given in the working space then incorrectly rounded, award full marks.
| Answer | Mark | Mark scheme |
|---|---|---|
| Increase (supported) | C1 | States increase with supporting reason eg “\(\dfrac{7}{10}\) is greater than \(\dfrac{7}{11}\)” “0.636 is less than 0.7” ….“cos increases as angle decreases” “decreasing the denominator increases the value of the fraction” “angle is now 45.6” (accept 45.5 – 45.6) |
If figures are given they must be correct (truncated or rounded).
22 \(ABCD\) is a trapezium.
Work out the size of angle \(CDA\).
Give your answer correct to 1 decimal place. (5)
| Answer | Mark | Notes |
|---|---|---|
| 32.3 | P1 | for using Pythagoras to find length of third side of triangle, eg \(7.5^2 - 6^2\) or \(6^2 + x^2 = 7.5^2\) or uses trigonometry to find angle in triangle eg \(\sin A = \dfrac{6}{7.5}\) or \(\cos B = \dfrac{6}{7.5}\) |
| P1 | (dep P1) for complete process to find length of third side of triangle eg \(\sqrt{7.5^2 - 6^2}\) or \(\sqrt{56.25 - 36}\) or \(\sqrt{20.25}\ (= 4.5)\) or uses trigonometry to find base length of triangle eg \(7.5 \times \cos \text{``}A\text{''}\) or \(7.5 \times \sin \text{``}B\text{''}\) or \(\dfrac{6}{\tan \text{``}A\text{''}}\) | |
| P1 | (dep P2) for \(24 - 10 - \text{``}4.5\text{''}\ (= 9.5)\) | |
| P1 | (indep) for process to find angle \(CDA\), eg \(\tan CDA = \dfrac{6}{\textit{base}}\) from right-angled triangle | |
| A1 | for answer in the range 32.2 to 32.3 |
20
\(OAC\) is a sector of a circle, centre \(O\), radius 10 m.
\(BA\) is the tangent to the circle at point \(A\).
\(BC\) is the tangent to the circle at point \(C\).
Angle \(AOC = 120^\circ\)
Calculate the area of the shaded region.
Give your answer correct to 3 significant figures. (5)
| Answer | Mark | Notes |
|---|---|---|
| 68.5 | B1 | for angle \(OAB = 90^\circ\) or angle \(OCB = 90^\circ\), may be seen on diagram |
| P1 | for a process to find the length of \(AB\) or the length of \(CB\) \((= 10\sqrt{3}\) oe\()\) eg \(10 \times \tan 60^\circ\ (= 17.3\ldots)\) or the length of \(OB\ (= 20)\), eg \(10 \div \cos 60^\circ\) | |
| P1 | for a process (dep previous P1) to find the area of the triangle \(OAB\) \((= 50\sqrt{3}\) oe\()\) or area of triangle \(OCB\) \((= 50\sqrt{3}\) oe\()\) or area of kite \(OABC\) \((= 100\sqrt{3}\) oe\()\) | |
| P1 | for a process to find the area of the sector \(OAC\) e.g. \(\dfrac{1}{3} \times \pi \times 10^2\ (= 104.7\ldots)\), accept rounded or truncated to 3 significant figures or more | |
| A1 | for 68.4 − 68.6 |
7 \(ABCD\) is a trapezium.
Work out the size of angle \(CDA\).
Give your answer correct to 1 decimal place. (5)
| Answer | Mark | Notes |
|---|---|---|
| 32.3 | P1 | for using Pythagoras to find length of third side of triangle, eg \(7.5^2 - 6^2\) or \(6^2 + x^2 = 7.5^2\) or uses trigonometry to find angle in triangle eg \(\sin A = \dfrac{6}{7.5}\) or \(\cos B = \dfrac{6}{7.5}\) |
| P1 | (dep P1) for complete process to find length of third side of triangle eg \(\sqrt{7.5^2 - 6^2}\) or \(\sqrt{56.25 - 36}\) or \(\sqrt{20.25}\ (= 4.5)\) or uses trigonometry to find base length of triangle eg \(7.5 \times \cos \text{``}A\text{''}\) or \(7.5 \times \sin \text{``}B\text{''}\) or \(\dfrac{6}{\tan \text{``}A\text{''}}\) | |
| P1 | (dep P2) for \(24 - 10 - \text{``}4.5\text{''}\ (= 9.5)\) | |
| P1 | (indep) for process to find angle \(CDA\), eg \(\tan CDA = \dfrac{6}{\textit{base}}\) from right-angled triangle | |
| A1 | for answer in the range 32.2 to 32.3 |
18
\(A\), \(B\) and \(C\) are points on a circle of radius 5 cm, centre \(O\).
\(DA\) and \(DC\) are tangents to the circle.
\(DO = 9\) cm
Work out the length of arc \(ABC\).
Give your answer correct to 3 significant figures. (5)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| Note \(DOC = DOA\), \(ADO = CDO\) | 21.6 | P1 | Recognises that \(OAD\) or \(OCD\) is 90° or right angle |
| P1 | for using trigonometry to set up an equation in \(DOA\) or \(ADO\) eg \(\cos DOA = \dfrac{5}{9}\) | ||
| P1 | for using inverse trigonometry to find \(DOA\) or \(ADO\) eg \(DOA = \cos^{-1} \dfrac{5}{9}\ (= 56.25\ldots)\) | ||
| P1 | for a complete process to find arc length \(ABC\) or \(AC\) eg \(\dfrac{360 - 2 \times \text{``}56.25..\text{''}}{360} \times 2 \times \pi \times 5\ (= 21.598..)\) or \(\dfrac{2 \times \text{``}56.25..\text{''}}{360} \times 2 \times \pi \times 5\ (= 9.8174..)\) | ||
| A1 | for answer in the range 21.5 to 21.65 |
7 \(ABC\) is a right-angled triangle.
Calculate the length of \(AB\).
Give your answer correct to 3 significant figures. (2)
| Answer | Mark | Notes |
|---|---|---|
| 5.86 | M1 | for \(\sin 23 = \dfrac{AB}{15}\) NB Allow any alternative equivalent method to form an equation in \(AB\) |
| A1 | 5.8 to 5.9 |
15
Diagram NOT accurately drawn
\(AB = 15\) m
\(BC = 24\) m
Angle \(BAD = 62^\circ\)
Work out the size of angle \(BCD\).
Give your answer correct to 1 decimal place. (5)
| Answer | Mark | Notes |
|---|---|---|
| 28.9 | 5 | M1 for \(\sin 62 = \dfrac{BD}{15}\) or \(\dfrac{BD}{\sin 62} = \dfrac{15}{\sin 90}\) oe M1 for \((BD =)\ 15 \times \sin 62\) or \(\dfrac{15}{\sin 90} \times \sin 62\) oe (= 13.24...) M1 for \(\tan BCD = \dfrac{\text{``}13.24\text{''}}{24}\) oe or \(\tan BDC = \dfrac{24}{\text{``}13.24\text{''}}\) with \(BDC\) clearly identified M1 for \(BCD = \tan^{-1}\dfrac{\text{``}13.24\text{''}}{24}\) oe or \(BDC = \tan^{-1}\dfrac{24}{\text{``}13.24\text{''}}\) with \(BDC\) clearly identified A1 for 28.8 – 28.9 OR M1 for \(\cos(90 - 62) = \dfrac{BD}{15}\) M1 for \((BD =)\ 15 \times \cos(90 - 62)\) (= 13.24...) M1 for \(\tan BCD = \dfrac{\text{``}13.24\text{''}}{24}\) oe or \(\tan BDC = \dfrac{24}{\text{``}13.24\text{''}}\) with \(BDC\) clearly identified M1 for \(BCD = \tan^{-1}\dfrac{\text{``}13.24\text{''}}{24}\) oe or \(BDC = \tan^{-1}\dfrac{24}{\text{``}13.24\text{''}}\) with \(BDC\) clearly identified A1 for 28.8 – 28.9 |
17
Diagram NOT accurately drawn
\(ABC\) is a right-angled triangle.
\(D\) is a point on \(AB\).
Angle \(ACD = 30^\circ\)
\(AD = 10.4\) cm
\(DB = 5.2\) cm
\(AC = 18\) cm
Work out the size of the angle marked \(x\).
Give your answer correct to 1 decimal place. (4)
| Answer | Mark | Notes |
|---|---|---|
| 30.1 | 4 | M1 for a correct trigonometric statement to find an unknown angle, eg. \(\sin(30 + x)\) or \(\cos A = \dfrac{10.4 + 5.2}{18}\) or \(\dfrac{\sin ADC}{18} = \dfrac{\sin 30}{10.4}\) M1 for a complete method to find the angle, eg. \(\sin^{-1}\left(\dfrac{10.4 + 5.2}{18}\right)\) (= 60.07...) or \(\cos^{-1}\left(\dfrac{10.4 + 5.2}{18}\right)\) (= 29.92..) or \(\sin^{-1}\left(\dfrac{18 \times \sin 30}{10.4}\right)\) (= 59.92.. or 180 – 59.92.. = 120.07..) M1 (dep on M2) for a fully complete method to find angle \(x\), eg. "60.07..". – 30 or 60 – "29.92.." or 90 – "59.92.." A1 for answer in the range 30.07 to 30.1 OR M1 for \((BC^2 =)\ 18^2 - (10.4 + 5.2)^2\) or \(BC^2 + (10.4 + 5.2)^2 = 18^2\) M1 for \((BC =)\ \sqrt{18^2 - (10.4 + 5.2)^2}\) (= 8.97...) M1 (dep on M2) for a fully complete method to find angle \(x\), eg. \(\tan^{-1}\left(\dfrac{5.2}{\text{``}8.97...\text{''}}\right)\) A1 for answer in the range 30.07 to 30.1 |
15
Diagram NOT accurately drawn
\(B\), \(C\) and \(D\) are points on the circumference of a circle, centre \(O\).
\(BOD\) is a diameter of the circle.
\(AO = 7\) cm Angle \(ABO = 90^\circ\) Angle \(OAB = 35^\circ\) Angle \(DBC = 70^\circ\)
| Answer | Mark | Notes |
|---|---|---|
| 1 | C1 for a complete reason eg Angles in a semicircle are \(90^\circ\), alternate segment theorem |
| Answer | Mark | Notes |
|---|---|---|
| 2.75 | 4 | M1 for \(7 \times \sin 35\) M1 for \(7 \times \sin 35 \times 2\) M1 (indep) for \(\text{``}DB\text{''} \times \cos 70\) A1 2.74 - 2.75 |
19
Diagram NOT accurately drawn
\(ABCD\) is a rectangle.
\(CDE\) is a straight line.
\(AB = 12\) cm
Angle \(ACB = 60^\circ\)
Angle \(EAC = 90^\circ\)
Calculate the length of \(CE\).
You must show all your working. (4)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(BC = \dfrac{12}{\tan 60} = 6.92(8…)\) \(DE = 6.92(…) \times \tan 30 = 4\) \(CE = 12 + 4\) \(AC = \dfrac{12}{\sin 60} = 13.8(5…)\) \(CE = \dfrac{13.8(5…)}{\cos 30}\) | 16 with supporting working | 4 | M1 for a method to find \(BC\) or \(AC\) or \(AD\) B1 for angle \(EAD = 30^\circ\) or \(AED = 60^\circ\) or \(ACD = 30^\circ\) or \(CAD = 60^\circ\) M1 for a method to find \(CE\) A1 for 15.9-16.1 with supporting working |
16 The diagram represents a metal frame.
Diagram NOT accurately drawn
The frame is made from four metal bars, \(AB\), \(AC\), \(BC\) and \(BD\).
Angle \(ABC\) = angle \(ADB = 90^\circ\)
\(AB = 5\) m
\(BC = 3\) m
Work out the total length of the four metal bars of the frame.
Give your answer correct to 3 significant figures. (5)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(AC^2 = 5^2 + 3^2\) \(AC = \sqrt{25 + 9}\) (=5.83) \(\dfrac{5}{5.83} = \dfrac{DB}{3}\) \(DB = \dfrac{5}{5.83} \times 3\) (= 2.57) \(5 + 3 + 5.83 + 2.57 =\) OR \(AC = \sqrt{25 + 9}\) (=5.83) \(\tan A = \dfrac{3}{5}\) \(A = 30.96\) \(\sin 30.96 = \dfrac{DB}{5}\) \(DB = 5 \times \sin 30.96\) (= 2.57) \(5 + 3 + 5.83 + 2.57 =\) | 16.4 | 5 | M1 for \((AC^2) = 5^2 + 3^2\) (=34) M1 for \(\sqrt{25 + 9}\) or \(\sqrt{34}\) (=5.83) M1 for \(\dfrac{5}{\text{'}5.83'} = \dfrac{DB}{3}\) or \(DB \times AC = 5 \times 3\) M1 for \((DB =)\ \dfrac{5}{\text{'}5.83'} \times 3\) A1 for 16.4 to 16.41 OR M1 for \((AC^2) = 5^2 + 3^2\) (=34) M1 for \(\sqrt{25 + 9}\) or \(\sqrt{34}\) (=5.83) M1 for using a correct trig ratio in an attempt to find angle \(A\) or angle \(C\), e.g. \(\tan A = \dfrac{3}{5}\), \(\sin A = \dfrac{3}{\text{'}5.83'}\), \(\cos C = \dfrac{3}{\text{'}5.83'}\) M1 for using \(DB\) in a a correct trig ratio, e.g. \(\sin \text{'}30.96\text{'} = \dfrac{DB}{5}\) A1 for 16.4 to 16.41 |
15 The diagram shows the positions of three turbines \(A\), \(B\) and \(C\).
Diagram NOT accurately drawn
\(A\) is 6 km due north of turbine \(B\).
\(C\) is 4.5 km due west of turbine \(B\).
| Answer | Mark | Notes |
|---|---|---|
| 7.5 | 3 | M1 for \(4.5^2 + 6^2\) (=56.25) M1 for \(\sqrt{56.25}\) or \(\sqrt{4.5^2 + 6^2}\) A1 for 7.5 |
| Answer | Mark | Notes |
|---|---|---|
| 217 | 4 | M1 for use of appropriate trig ratio eg \(\tan CAB = \dfrac{4.5}{6}\) (= 0.75), \(\sin CAB = \dfrac{4.5}{\text{``}7.5\text{''}}\) (= 0.6), \(\cos CAB = \dfrac{6}{\text{``}7.5\text{''}}\) (= 0.8) M1 for inverse trig shown correctly eg \(CAB = \tan^{-1} \dfrac{4.5}{6}\) (= 0.75), \(CAB = \sin^{-1} \dfrac{4.5}{\text{``}7.5\text{''}}\) (= 0.6), \(CAB = \cos^{-1} \dfrac{6}{\text{``}7.5\text{''}}\) (= 0.8) A1 for 36.8 to 37 (or 53 to 53.2 if identified as \(ACB\)) B1ft for bearing 180 + “36.8” if “36.8” is not 40-50 eg 216.8 to 217 |
18 \(ABC\) is an isosceles triangle.
Diagram NOT accurately drawn
Work out the area of the triangle.
Give your answer correct to 3 significant figures. (4)
| Answer | Mark | Notes |
|---|---|---|
| 49.5 | 4 | M1 for \(\tan 54 = \dfrac{\text{height}}{6}\) M1 for (height =) \(6 \times \tan 54\) (=8.2-8.3) M1 for \(\frac{1}{2} \times\) ‘8.258..’ \(\times 12\) A1 for 49.2 - 50 OR M1 for \(\cos 54 = \dfrac{6}{AC}\) M1 for (\(AC =\)) \(\dfrac{6}{\cos 54}\) (=10.2(07...)) M1 for \(\frac{1}{2} \times 12 \times\) ‘10.207’ \(\times \sin 54\) A1 for 49.2 - 50 OR M1 for \(\dfrac{AC}{\sin 54} = \dfrac{12}{\sin 72}\) M1 for (\(AC =\)) \(\dfrac{12}{\sin 72} \times \sin 54\) (=10.2(07...)) M1 for \(\frac{1}{2} \times 12 \times\) ‘10.207’ \(\times \sin 54\) A1 for 49.2 – 50 |
*20 The diagram shows a ladder leaning against a vertical wall.
Diagram NOT accurately drawn
The ladder stands on horizontal ground.
The length of the ladder is 6 m.
The bottom of the ladder is 2.25 m from the bottom of the wall.
A ladder is safe to use when the angle marked \(y\) is about \(75^\circ\).
Is the ladder safe to use?
You must show all your working. (3)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(\cos y = 2.25 \div 6\) \(y = \cos^{-1}(2.25 \div 6)\) OR \(6\cos 75 = 1.55\ldots\) | The ladder is not safe because \(y\) is not near to 75 | 3 | M1 for \(\cos y = 2.25 \div 6\) oe M1 for \(\cos^{-1}(2.25 \div 6)\) C1 for sight of 67-68 and a statement eg this angle is NOT (near to) 75° and so the ladder is not steep enough and so not safe. OR M1 for \(\cos 75 = x \div 6\) M1 for \(6\cos 75\) C1 for sight of 1.55(29…) and a statement eg that 2.25 NOT (near to) 1.55 and so the ladder is not steep enough and so not safe. |
13
Diagram NOT accurately drawn
\(ABC\) is a right-angled triangle.
\(AC = 6\) cm
\(AB = 13\) cm
Diagram NOT accurately drawn
\(PQR\) is a right-angled triangle.
\(PR = 17\) cm
\(PQ = 25\) cm
| Answer | Mark | Notes |
|---|---|---|
| 11.5 | 3 | M1 for \(13^2 - 6^2\) or \(169 - 36\) or 133 M1 (dep on M1) for \(\sqrt{13^2 - 6^2}\) or \(\sqrt{133}\) A1 for answer in the range 11.5 − 11.6 |
| Answer | Mark | Notes |
|---|---|---|
| 47.2 | 3 | M1 for \(\cos(RPQ) = \frac{17}{25}\) oe OR \(\sin PQR = \frac{17}{25}\) with \(PQR\) clearly identified M1 for (\(RPQ =\)) \(\cos^{-1}\frac{17}{25}\) oe OR \(PQR = \sin^{-1}\frac{17}{25}\) with \(PQR\) clearly identified A1 for answer in the range 47.1 − 47.2 SC : B2 for an answer of 0.823(033...) or 52.3(95...) or 52.4 |
17
Diagram NOT accurately drawn
Calculate the value of \(x\).
Give your answer correct to 3 significant figures. (3)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(\sin 60^\circ = \dfrac{x}{32}\) \(x = 32 \times \sin 60\) (=27.712...) | 27.7 | 3 | M1 \(\sin 60 = \dfrac{x}{32}\) or \(\dfrac{x}{\sin 60} = \dfrac{32}{\sin 90}\) oe M1 (\(x =\)) \(32 \times \sin 60\) or (\(x =\)) \(\dfrac{32}{\sin 90} \times \sin 60\) A1 27.7 − 27.72 OR M1 \(\cos(90 - 60) = \dfrac{x}{32}\) M1 (\(x =\)) \(32 \times \cos(90 - 60)\) A1 27.7 − 27.72 Radians : − 9.7539398... Gradians : 25.888554... SC : B2 for an answer in the range (−) 9.75 to (−)9.754 or 25.8 to 25.9 |
18 The diagram shows a quadrilateral \(ABCD\).
Diagram NOT accurately drawn
\(AB = 16\) cm.
\(AD = 12\) cm.
Angle \(BCD = 40^\circ\)
Angle \(ADB\) = angle \(CBD = 90^\circ\)
Calculate the length of \(CD\).
Give your answer correct to 3 significant figures. (5)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(BD^2 + 12^2 = 16^2\) oe \(BD = \sqrt{256 - 144}\) (=10.58...) \(\sin 40 = \dfrac{\text{‘}10.58\text{’}}{CD}\) \(CD = \dfrac{\text{‘}10.58\text{’}}{\sin 40}\) | 16.5 | 5 | M1 for \(BD^2 + 12^2 = 16^2\) oe or \(16^2 - 12^2\) or 112 seen M1 for \(\sqrt{256 - 144}\) or \(\sqrt{112}\) (=10.58...) M1 for \(\sin 40 = \dfrac{\text{‘}10.58\text{’}}{CD}\) or \(\cos 50 = \dfrac{\text{‘}10.58\text{’}}{CD}\) M1 for (\(CD =\)) \(\dfrac{\text{‘}10.58\text{’}}{\sin 40}\) or \(\dfrac{\text{‘}10.58\text{’}}{\cos 50}\) A1 for 16.4 – 16.5 OR M1 for \(BD^2 + 12^2 = 16^2\) oe or \(16^2 - 12^2\) or 112 seen M1 for \(\sqrt{256 - 144}\) or \(\sqrt{112}\) (=10.58...) M1 for (\(BC =\)) ‘10.58’ × tan 50 or \(\dfrac{\text{‘}10.58\text{’}}{\tan 40}\) (=12.6…) M1 for \(\sqrt{\text{‘}12.6\text{’}^2 + \text{‘}10.58...\text{’}^2}\) A1 for 16.4 – 16.5 |
16
Diagram NOT accurately drawn
\(LMN\) is a right-angled triangle.
\(MN = 9.6\) cm.
\(LM = 6.4\) cm.
Calculate the size of the angle marked \(x^\circ\).
Give your answer correct to 1 decimal place. (3)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(\cos x = \dfrac{6.4}{9.6}\) \(x = \cos^{-1}\dfrac{6.4}{9.6} =\) | 48.2 | 3 | M1 for \(\cos x = \dfrac{6.4}{9.6}\) or \(\cos x = 0.66(6...)\) or \(\cos x = 0.67\) M1 for \(\cos^{-1}\dfrac{6.4}{9.6}\) or \(\cos^{-1} 0.66(6...)\) or \(\cos^{-1} 0.67\) A1 for 48.1 – 48.2 OR Correct use of Pythagoras and then trigonometry, no marks until M1 for \(\sin x = \dfrac{\text{‘}7.155\text{’}}{9.6}\) or \(\tan x = \dfrac{\text{‘}7.155\text{’}}{6.4}\) or \(\sin x = \dfrac{\text{‘}7.155\text{’}}{9.6} \times \sin 90\) or \(\cos x = \dfrac{6.4^2 + 9.6^2 - \text{‘}7.155\text{’}^2}{2 \times 6.4 \times 9.6}\) M1 for \(\sin^{-1}\dfrac{\text{‘}7.155\text{’}}{9.6}\) or \(\tan^{-1}\dfrac{\text{‘}7.155\text{’}}{6.4}\) or \(\sin^{-1}\left(\dfrac{\text{‘}7.155\text{’}}{9.6} \times \sin 90\right)\) or \(\cos^{-1}\left(\dfrac{6.4^2 + 9.6^2 - \text{‘}7.155\text{’}^2}{2 \times 6.4 \times 9.6}\right)\) A1 for 48.1 – 48.2 SC B2 for 0.841... (using rad) or 53.5... (using grad) |
26
Not drawn accurately
Use trigonometry to work out the value of \(x\).
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: uses or states tan | ||
| tan stated or used | M1 | implied by [0.726, 0.73] or [1.376, 1.4] |
| \(\tan 36 = \dfrac{x}{4}\) or \(4 \times \tan 36\) or \(\tan 54 = \dfrac{4}{x}\) or \(\dfrac{4}{\tan 54}\) | M1dep | oe accept [0.726, 0.73] for tan 36 accept [1.376, 1.4] for tan 54 |
| [2.9, 2.91] with M2 awarded | A1 | accept 3 with M2 awarded |
| Alternative method 2: works out hypotenuse first | ||
| (hypotenuse \(=\)) \(\dfrac{4}{\sin 54}\) or \(\dfrac{4}{\cos 36}\) or 4.9(…) | M1 | oe accept [0.8, 0.81] for sin 54 or cos 36 |
| \(\sqrt{(\text{their } 4.9(\ldots))^2 - 4^2}\) or \(\sqrt{[8, 8.45]}\) or their 4.9(…) \(\times \sin 36\) or their 4.9(…) \(\times \cos 54\) | M1dep | oe |
| [2.9, 2.91] with M2 awarded | A1 | accept 3 with M2 awarded |
| Alt 1 SOHCAHTOA with TOA selected (eg circled) | M1 |
| [2.9, 2.91] with no working | M0M0A0 |
| [2.9, 2.91] from accurate/scale drawing only | M0M0A0 |
10
Not drawn accurately
Use trigonometry to work out the value of \(x\).
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: uses or states tan | ||
| tan stated or used | M1 | implied by [0.726, 0.73] or [1.376, 1.4] |
| \(\tan 36 = \dfrac{x}{4}\) or \(4 \times \tan 36\) or \(\tan 54 = \dfrac{4}{x}\) or \(\dfrac{4}{\tan 54}\) | M1dep | oe accept [0.726, 0.73] for tan 36 accept [1.376, 1.4] for tan 54 |
| [2.9, 2.91] with M2 awarded | A1 | accept 3 with M2 awarded |
| Alternative method 2: works out hypotenuse first | ||
| (hypotenuse \(=\)) \(\dfrac{4}{\sin 54}\) or \(\dfrac{4}{\cos 36}\) or 4.9(…) | M1 | oe accept [0.8, 0.81] for sin 54 or cos 36 |
| \(\sqrt{(\text{their } 4.9(\ldots))^2 - 4^2}\) or \(\sqrt{[8, 8.45]}\) or their 4.9(…) \(\times \sin 36\) or their 4.9(…) \(\times \cos 54\) | M1dep | oe |
| [2.9, 2.91] with M2 awarded | A1 | accept 3 with M2 awarded |
| Alt 1 SOHCAHTOA with TOA selected (eg circled) | M1 |
| [2.9, 2.91] with no working | M0M0A0 |
| [2.9, 2.91] from accurate/scale drawing only | M0M0A0 |
22 Use trigonometry to work out the value of \(x\). [3 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – using sin 40 | ||
| sin chosen or used | M1 | |
| \(21 \times \sin 40\) | M1dep | accept \(21 \times [0.64, 0.643]\) |
| [13.49, 13.5] | A1 | |
| Alternative method 2 – using cos 50 | ||
| \(\cos (90 - 40)\) | M1 | |
| \(21 \times \cos (90 - 40)\) | M1dep | oe accept \(21 \times [0.64, 0.643]\) |
| [13.49, 13.5] | A1 | |
| Alternative method 3 – finds base then uses Pythagoras | ||
| \(21^2 - (21 \sin (90 - 40))^2\) | M1 | oe complete method except square root |
| \(\sqrt{21^2 - (21 \sin (90 - 40))^2}\) or \(\sqrt{[182.2, 182.22]}\) | M1dep | oe |
| [13.49, 13.5] | A1 | |
| Check diagram for working | |
| Allow correct use of sine rule to indicate sin | |
| Ignore rounding or truncating after the correct answer is seen eg answer 14 after 13.5 seen | M1M1A1 |
| \(\sin 40 \times 21\) | M2 |
| sin may be indicated by eg circling S in SOH CAH TOA | |
| Do not accept answers from full sized or scale drawing | |
| sin 50 used (unless using Alt 3) | M0 |
13 Two right-angled triangles are joined to make a larger triangle.
Not drawn accurately
Work out the size of angle \(x\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\tan 38 = \dfrac{h}{15}\) or \(\left(\dfrac{15}{\cos 38}\right)^2 - 15^2\) or [137.3, 137.342] | M1 | oe eg \(\tan (90 - 38) = \dfrac{15}{h}\) or \(\dfrac{\sin 38}{h} = \dfrac{\sin 52}{15}\) any letter or phrase for \(h\) eg opposite |
| \(15 \times \tan 38\) or \(\sqrt{\left(\dfrac{15}{\cos 38}\right)^2 - 15^2}\) or [11.7, 11.72] | M1dep | oe eg \(15 \div \tan (90 - 38)\) or \(\dfrac{15 \sin 38}{\sin 52}\) |
| \(\sin x = \dfrac{\text{their } [11.7, 11.72]}{14}\) or \(\sin x = [0.835, 0.84]\) | M1dep | oe eg \(\sin x = \dfrac{\sin 90}{14} \times\) their [11.7, 11.72] implied by \(\sin^{-1} [0.835, 0.84]\) |
| [56.6, 57.14012] | A1 |
Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
19
Not drawn accurately
Use trigonometry to work out the value of \(x\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\sin\) chosen or used | M1 | |
| \(31 \times \sin 24\) | M1dep | accept \(31 \times [0.4, 0.41]\) |
| [12.6, 12.61] | A1 | accept 13 if M2 awarded |
| Alternative method 2 | ||
| \(\cos (90 - 24)\) | M1 | |
| \(31 \times \cos (90 - 24)\) | M1dep | accept \(31 \times [0.4, 0.41]\) |
| [12.6, 12.61] | A1 | accept 13 if M2 awarded |
| Check diagram for working | |
| Allow correct use of sine rule to indicate \(\sin 24\) | |
| Ignore rounding or truncating after the correct answer is seen | |
| \(\sin 24 \times 31\) | M2 |
| Do not accept answers from full sized or scale drawing | |
| \(\sin\) may be indicated by eg circling S in SOH CAH TOA |
18 A diagonal of a rectangle is 23.7 cm long.
The diagonal makes an angle of \(52^\circ\) with a side of length \(x\) cm
Work out the value of \(x\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: only uses trigonometry | ||
| \(\cos 52 = \dfrac{x}{23.7}\) | M1 | oe eg \(\sin(90 - 52) = \dfrac{x}{23.7}\) or \(\dfrac{x}{\sin 38} = \dfrac{23.7}{\sin 90}\) accept [0.61, 0.62] for cos 52 or sin 38 |
| \(23.7 \times \cos 52\) | M1dep | oe eg \(23.7 \times \sin 38 \div \sin 90\) accept [0.61, 0.62] for cos 52 or sin 38 |
| [14.59, 14.6] | A1 | SC1 [18.4, 18.723] |
| Alternative method 2: uses trigonometry and Pythagoras | ||
| \(23.7^2\) and \((23.7 \times \sin 52)^2\) or [561.6, 561.7] and [338, 351] | M1 | oe accept [0.78, 0.79] for sin 52 accept [18.4, 18.723] for \(23.7 \times \sin 52\) |
| \(\sqrt{23.7^2 - (23.7 \times \sin 52)^2}\) or \(\sqrt{[210.6, 223.7]}\) | M1dep | oe accept [0.78, 0.79] for sin 52 accept [18.4, 18.723] for \(23.7 \times \sin 52\) |
| [14.59, 14.6] | A1 | SC1 [18.4, 18.723] |
M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
SC1 is from a diagonal making an angle of \(38^\circ\) with \(x\)
28 \(VABCD\) is a pyramid with a horizontal square base.
\(X\) is the centre of the base.
\(V\) is vertically above \(X\).
\(BD = 18\) cm
Angle \(VBX\) = 72°
Work out the length of \(VB\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\cos 72 = \dfrac{9}{x}\) | M1 | oe eg \(\dfrac{x}{\sin 72} = \dfrac{18}{\sin 36}\) \(x\) can be any letter or \(VB\) or \(VA\) or \(VC\) or \(VD\) |
| \(\dfrac{9}{\cos 72}\) | M1dep | oe eg \(\dfrac{18 \times \sin 72}{\sin 36}\) |
| 29.1(2…) | A1 | accept 29 with M1 scored |
Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
23 Use trigonometry to work out the size of angle \(w\). [3 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| cos chosen or used | M1 | |
| \(\cos w = \dfrac{6.7}{8.3}\) or \(\cos^{-1} \dfrac{6.7}{8.3}\) | M1dep | any letter or symbol for \(w\) accept 0.807(…) or 0.81 for \(\dfrac{6.7}{8.3}\) |
| [35.9, 36.2] | A1 | |
| Alternative method 2 | ||
| \(\sin x = \dfrac{6.7}{8.3}\) or \(\sin^{-1} \dfrac{6.7}{8.3}\) or [53.8, 54.1] | M1 | any letter or symbol other than \(w\) accept 0.807(…) or 0.81 for \(\dfrac{6.7}{8.3}\) |
| \(90 -\) their [53.8, 54.1] | M1dep | |
| [35.9, 36.2] | A1 | |
| Alternative method 3 | ||
| \(\sqrt{8.3^2 - 6.7^2}\) or \(\sqrt{68.89 - 44.89}\) or \(\sqrt{24}\) or \(2\sqrt{6}\) or [4.89, 4.9] and \(\sin^{-1} \dfrac{\text{their } [4.89, 4.9]}{8.3}\) or \(\tan^{-1} \dfrac{\text{their } [4.89, 4.9]}{6.7}\) | M2 | full method to work out the missing length and use it correctly to work out the value of \(w\) any letter or symbol for \(w\) |
| [35.9, 36.2] | A1 | |
| Use of sine rule follows Alt method 2 | |
| \(\sin w = \dfrac{6.7}{8.3}\) without \(\sin^{-1} \dfrac{6.7}{8.3}\) or [53.8, 54.1] | M0 |
| \(\cos w = 0.807\) | M1M1 |
| \(\cos^{-1} w = \dfrac{6.7}{8.3}\) or \(\cos = \dfrac{6.7}{8.3}\) unless recovered | M1M0 |
9 Use trigonometry to work out the size of angle \(w\). [3 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| cos chosen or used | M1 | |
| \(\cos w = \dfrac{6.7}{8.3}\) or \(\cos^{-1} \dfrac{6.7}{8.3}\) | M1dep | any letter or symbol for \(w\) accept 0.807(…) or 0.81 for \(\dfrac{6.7}{8.3}\) |
| [35.9, 36.2] | A1 | |
| Alternative method 2 | ||
| \(\sin x = \dfrac{6.7}{8.3}\) or \(\sin^{-1} \dfrac{6.7}{8.3}\) or [53.8, 54.1] | M1 | any letter or symbol other than \(w\) accept 0.807(…) or 0.81 for \(\dfrac{6.7}{8.3}\) |
| \(90 -\) their [53.8, 54.1] | M1dep | |
| [35.9, 36.2] | A1 | |
| Alternative method 3 | ||
| \(\sqrt{8.3^2 - 6.7^2}\) or \(\sqrt{68.89 - 44.89}\) or \(\sqrt{24}\) or \(2\sqrt{6}\) or [4.89, 4.9] and \(\sin^{-1} \dfrac{\text{their } [4.89, 4.9]}{8.3}\) or \(\tan^{-1} \dfrac{\text{their } [4.89, 4.9]}{6.7}\) | M2 | full method to work out the missing length and use it correctly to work out the value of \(w\) any letter or symbol for \(w\) |
| [35.9, 36.2] | A1 | |
| Use of sine rule follows Alt method 2 | |
| \(\sin w = \dfrac{6.7}{8.3}\) without \(\sin^{-1} \dfrac{6.7}{8.3}\) or [53.8, 54.1] | M0 |
| \(\cos w = 0.807\) | M1M1 |
| \(\cos^{-1} w = \dfrac{6.7}{8.3}\) or \(\cos = \dfrac{6.7}{8.3}\) unless recovered | M1M0 |
23 Use trigonometry to work out the value of \(x\).
Not drawn accurately
[3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – using tangent of an angle | ||
| tan chosen or used | M1 | |
| \(\tan 58 = \dfrac{x}{46}\) or \(46 \times \tan 58\) or \(\tan 32 = \dfrac{46}{x}\) or \(\dfrac{46}{\tan 32}\) | M1dep | oe |
| [73.6, 74] | A1 | |
| Alternative method 2 – finding hypotenuse first | ||
| \(\dfrac{46}{\cos 58}\) or \(\dfrac{46}{\sin 32}\) or 86.8(…) or 87 | M1 | oe |
| \(\sqrt{(\text{their } 86.8(\ldots))^2 - 46^2}\) or \(\sqrt{5418.(\ldots)}\) or their \(86.8(\ldots) \times \sin 58\) or their \(86.8(\ldots) \times \cos 32\) | M1dep | oe |
| [73.6, 74] | A1 | |
| Do not accept scale drawing | |
| Answer 73 after answer in range seen | M1M1A1 |
| \(\dfrac{\sin 32}{46} = \dfrac{\sin 58}{x}\) | M1 |
17
Not drawn accurately
Work out the size of angle \(x\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Works out \(AC\) and uses it in triangle \(ABC\) | ||
| \(\cos 37 = \dfrac{AC}{4}\) | M1 | oe eg \(\sin 53 = \dfrac{AC}{4}\) allow [0.798, 0.8] for cos 37 or sin 53 |
| \((AC =)\ 4 \times \cos 37\) or \((AC =)\) [3.19, 3.2] | M1dep | oe eg \((AC =)\ 4 \times \sin 53\) allow [0.798, 0.8] for cos 37 or sin 53 may be seen on diagram |
| \(\sin x = \dfrac{\text{their } [3.19, 3.2]}{9.3}\) or \((x =)\ \sin^{-1}[0.34, 0.3441]\) | M1dep | oe eg \(\cos x = \dfrac{\sqrt{9.3^2 - \text{their } [3.19, 3.2]^2}}{9.3}\) or \((x =)\ 90 - \cos^{-1}[0.34, 0.3441]\) |
| [19.87, 20.13] | A1 | |
| Alternative method 2 Works out angle \(ADC\) and uses it in triangle \(ABD\) | ||
| (angle \(ADC =\)) \(90 - 37\) or (angle \(ADC =\)) 53 | M1 | oe eg (angle \(ADC =\)) \(180 - 90 - 37\) may be seen on diagram |
| \(\dfrac{\sin x}{4} = \dfrac{\sin(90 - 37)}{9.3}\) | M1dep | oe eg \(\dfrac{4}{\sin x} = \dfrac{9.3}{\sin 53}\) |
| \((\sin x =)\ \dfrac{\sin(90 - 37)}{9.3} \times 4\) or \((x =)\ \sin^{-1}[0.34, 0.3441]\) | M1dep | oe |
| [19.87, 20.13] | A1 | |
| Up to M3 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Allow any unambiguous notation for angles eg allow \(B\) for \(x\) | |
| Alt 1 Allow any unambiguous notation for \(AC\) eg \(y\) (condone \(x\) if clearly referring to \(AC\)) | |
| Alt 1 1st M1 must be an equation where \(AC\) is the only variable eg \(AC^2 + (4\sin 37)^2 = 4^2\) | M1 |
| Alt 1 A calculation that leads to \(AC\) scores M1M1 eg \(\sqrt{4^2 - (4\sin 37)^2}\) | M1M1 |
| Alt 1 3rd M1 must have \(\sin x\) (or \(\cos x\)) as the subject or be a calculation that leads to \(x\) | |
| Alt 2 53 only marked at angle \(BAC\) on diagram | M0 |
16
Not drawn accurately
In this right-angled triangle,
\(a = 16\) cm
\(a : c = 4 : 5\)
Work out the area of the triangle. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – using Pythagoras’ theorem or 3, 4, 5 triangle | ||
| \(16 \div 4 \times 5\) or 20 (cm) or identifies triangle as 3, 4, 5 | M1 | oe length of \(c\) may be on diagram |
| \(\sqrt{(\text{their } 20)^2 - 16^2}\) or \(\sqrt{400 - 256}\) or \(\sqrt{144}\) or \(4 \times 3\) | M1dep | |
| 12 (cm) | A1 | length of \(b\) may be on diagram |
| 96 | A1ft | ft \(\dfrac{1}{2} \times 16 \times\) their 12 with M2 awarded |
| Alternative method 2 – using trigonometry and \(\tfrac{1}{2}ab\sin C\) formula | ||
| \(16 \div 4 \times 5\) or 20 (cm) | M1 | oe length of \(c\) may be on diagram |
| \(\cos^{-1}\left(\dfrac{16}{20}\right)\) or 36.8(…) or 36.9 | M1dep | angle between sides \(a\) and \(c\) |
| \(\dfrac{1}{2} \times 16 \times 20 \times \sin(\text{their } 36.8(\ldots))\) | M1dep | dep on M2 |
| 96 | A1 | |
| \(\dfrac{1}{2} \times 16 \times 12 \times \sin 90\) | M1M1M1 |
10 Use trigonometry to work out the value of \(x\). [3 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – using tangent of an angle | ||
| tan chosen or used | M1 | |
| \(\tan 58 = \dfrac{x}{46}\) or \(46 \times \tan 58\) or \(\tan 32 = \dfrac{46}{x}\) or \(\dfrac{46}{\tan 32}\) | M1dep | oe |
| [73.6, 74] | A1 | |
| Alternative method 2 – finding hypotenuse first | ||
| \(\dfrac{46}{\cos 58}\) or \(\dfrac{46}{\sin 32}\) or 86.8(…) or 87 | M1 | oe |
| \(\sqrt{(\text{their } 86.8(\ldots))^2 - 46^2}\) or \(\sqrt{5418.(\ldots)}\) or their \(86.8(\ldots) \times \sin 58\) or their \(86.8(\ldots) \times \cos 32\) | M1dep | oe |
| [73.6, 74] | A1 | |
| Do not accept scale drawing | |
| Answer 73 after answer in range seen | M1M1A1 |
| \(\dfrac{\sin 32}{46} = \dfrac{\sin 58}{x}\) | M1 |
27 Use trigonometry to work out the value of \(x\).
Not drawn accurately
[3 marks]
| Answer | Mark | Comments |
|---|---|---|
| cos identified or sin 30 identified | M1 | oe |
| \((\cos 60 =)\ \dfrac{1}{2}\) or \((\sin 30 =)\ \dfrac{1}{2}\) | M1 | oe may be seen in a table |
| 4 | A1 |
| Check diagram for working 4 written next to the \(x\) on the diagram is full marks unless contradicted | |
| Cos can be identified by, for example, CAH circled in SOHCAHTOA | |
| \(8 \cos 60\) \(8 \times \dfrac{1}{2}\) | M1M0 M1M1 |
| Answer 4 from creating an equilateral triangle | M1M1A1 |
| Answer only 4 | M1M1A1 |
| Answer 4 from drawing a scale or full-sized diagram of the right-angled triangle | M0A0 |
21 Use trigonometry to work out the size of angle \(x\). [3 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| tan identified | M1 | oe eg \(\tan^{-1}\) |
| \(\tan x = \dfrac{10}{4}\) or \(\tan x = \dfrac{5}{2}\) or \(\tan x = 2.5\) | M1dep | oe eg \(\tan^{-1} \dfrac{10}{4}\) or \(90 - \tan^{-1} \dfrac{4}{10}\) |
| [68, 68.2] | A1 | SC1 [21.8, 22] |
| Alternative method 2 | ||
| \(\sin x = \dfrac{10}{\sqrt{4^2 + 10^2}}\) or \(\cos x = \dfrac{4}{\sqrt{4^2 + 10^2}}\) | M2 | oe eg \(\sin x = \dfrac{10}{\sqrt{116}}\) or \(\sin^{-1} \dfrac{10}{\sqrt{4^2 + 10^2}}\) or \(\cos x = \dfrac{4}{\sqrt{116}}\) or \(\cos^{-1} \dfrac{4}{\sqrt{4^2 + 10^2}}\) or \(90 - \sin^{-1} \dfrac{4}{\sqrt{4^2 + 10^2}}\) or \(90 - \cos^{-1} \dfrac{10}{\sqrt{4^2 + 10^2}}\) |
| [68, 68.2] | A1 | SC1 [21.8, 22] |
| Accept 10.77 or 10.8 or \(2\sqrt{29}\) for \(\sqrt{116}\) | |
| Tan can be identified by, for example, circling TOA in SOHCAHTOA | |
| Answer from accurate drawing | M0M0A0 |
| \(\sin x = \dfrac{10 \sin 90}{\sqrt{116}}\) | M2 |
| \((x =)\) tan 2.5 or \((x =)\) tan 0.4 or \((x =)\) \(\tan \left(\dfrac{10}{4}\right)^{-1}\) unless recovered | M1M0A0 |
| \(\tan = \dfrac{10}{4}\) or \(\tan = \dfrac{4}{10}\) or \(\tan x = \dfrac{4}{10}\) with no further correct working | M1M0A0 |
4 Use trigonometry to work out the size of angle \(x\).
Not drawn accurately
[3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| tan identified | M1 | oe eg \(\tan^{-1}\) |
| \(\tan x = \dfrac{10}{4}\) or \(\tan x = \dfrac{5}{2}\) or \(\tan x = 2.5\) | M1dep | oe eg \(\tan^{-1} \dfrac{10}{4}\) or \(90 - \tan^{-1} \dfrac{4}{10}\) |
| [68, 68.2] | A1 | SC1 [21.8, 22] |
| Alternative method 2 | ||
| \(\sin x = \dfrac{10}{\sqrt{4^2 + 10^2}}\) or \(\cos x = \dfrac{4}{\sqrt{4^2 + 10^2}}\) | M2 | oe eg \(\sin x = \dfrac{10}{\sqrt{116}}\) or \(\sin^{-1} \dfrac{10}{\sqrt{4^2 + 10^2}}\) or \(\cos x = \dfrac{4}{\sqrt{116}}\) or \(\cos^{-1} \dfrac{4}{\sqrt{4^2 + 10^2}}\) or \(90 - \sin^{-1} \dfrac{4}{\sqrt{4^2 + 10^2}}\) or \(90 - \cos^{-1} \dfrac{10}{\sqrt{4^2 + 10^2}}\) |
| [68, 68.2] | A1 | SC1 [21.8, 22] |
| Accept 10.77 or 10.8 or \(2\sqrt{29}\) for \(\sqrt{116}\) | |
| Tan can be identified by, for example, circling TOA in SOHCAHTOA | |
| Answer from accurate drawing | M0M0A0 |
| \(\sin x = \dfrac{10 \sin 90}{\sqrt{116}}\) | M2 |
| (\(x\) =) tan 2.5 or (\(x\) =) tan 0.4 or (\(x\) =) \(\tan \left(\dfrac{10}{4}\right)^{-1}\) unless recovered | M1M0A0 |
| \(\tan = \dfrac{10}{4}\) or \(\tan = \dfrac{4}{10}\) or \(\tan x = \dfrac{4}{10}\) with no further correct working | M1M0A0 |
31 Here is a right-angled triangle.
Not drawn accurately
Use trigonometry to work out the value of \(x\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\sin 30 = \dfrac{x}{10}\) or \((x =)\ 10 \sin 30\) | M1 | oe eg \(\dfrac{x}{\sin 30} = \dfrac{10}{\sin 90}\) |
| \(\sin 30 = 0.5\) | M1 | oe may be seen in a table \(0.5 = \dfrac{x}{10}\) oe scores M1M1 |
| 5 | A1 | |
| Alternative method 2 | ||
| Correct trigonometric method to show that the length of the missing side is \(5\sqrt{3}\) | M1 | oe |
| \(\sqrt{(5\sqrt{3})^2 + x^2} = 10\) | M1dep | oe |
| 5 | A1 | |
Accept use of cos 60 instead of sin 30
12 Here is a right-angled triangle.
Not drawn accurately
Use trigonometry to work out the value of \(x\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\sin 30 = \dfrac{x}{10}\) or (\(x\) =) \(10 \sin 30\) | M1 | oe eg \(\dfrac{x}{\sin 30} = \dfrac{10}{\sin 90}\) |
| \(\sin 30 = 0.5\) | M1 | oe may be seen in a table \(0.5 = \dfrac{x}{10}\) oe scores M1M1 |
| 5 | A1 | |
| Alternative method 2 | ||
| Correct trigonometric method to show that the length of the missing side is \(5\sqrt{3}\) | M1 | oe |
| \(\sqrt{(5\sqrt{3})^2 + x^2} = 10\) | M1dep | oe |
| 5 | A1 | |
Accept use of cos 60 instead of sin 30
27 Use trigonometry to work out the size of angle \(x\). [2 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| cos and \(\dfrac{9}{18}\) oe identified | M1 | |
| 60 | A1 | |
| Alternative method 2 | ||
| sin and \(\dfrac{\sqrt{18^2 - 9^2}}{18}\) identified or tan and \(\dfrac{\sqrt{18^2 - 9^2}}{9}\) identified | M1 | |
| 60 | A1 | |
| Accept an embedded answer, eg \(\cos 60 = \dfrac{9}{18}\) with no further working | M1A1 |
| \(180 \div 3 = 60\) | M0A0 |
18 This shape is made from two right-angled triangles and a rectangle.
Not drawn accurately
Work out the size of angle \(x\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\tan 62 = \dfrac{h}{5}\) | M1 | oe eg \(\tan (90 - 62) = \dfrac{5}{h}\) or \(\dfrac{h}{\sin 62} = \dfrac{5}{\sin 28}\) any letter |
| \(5 \times \tan 62\) or 9.4(0…) | M1dep | oe eg \(\dfrac{5}{\tan 28}\) or \(\dfrac{5}{\sin 28} \times \sin 62\) |
| \(\sin x = \dfrac{\text{their } 9.4(0\ldots)}{12}\) or \(\sin x =\) [0.78, 0.784] | M1dep | oe eg \(\sin x = \dfrac{5 \times \tan 62}{12}\) or \(\cos x = \dfrac{\sqrt{12^2 - \text{their } 9.4^2}}{12}\) |
| [51.536, 51.63] | A1 | accept 52 with M3 seen |
| Alternative method 2 | ||
| \(\left(\dfrac{5}{\cos 62}\right)^2 - 5^2\) or [88.4, 88.43] | M1 | oe |
| \(\sqrt{\left(\dfrac{5}{\cos 62}\right)^2 - 5^2}\) or 9.4(0…) | M1dep | oe |
| \(\sin x = \dfrac{\text{their } 9.4(0\ldots)}{12}\) or \(\sin x =\) [0.78, 0.784] | M1dep | oe eg \(\cos x = \dfrac{\sqrt{12^2 - \text{their } 9.4^2}}{12}\) |
| [51.536, 51.63] | A1 | accept 52 with M3 seen |
| Answer in range with truncation to 51 | M1M1M1A1 |
5 Use trigonometry to work out the size of angle \(x\). [2 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| cos and \(\dfrac{9}{18}\) oe identified | M1 | |
| 60 | A1 | |
| Alternative method 2 | ||
| sin and \(\dfrac{\sqrt{18^2 - 9^2}}{18}\) identified or tan and \(\dfrac{\sqrt{18^2 - 9^2}}{9}\) identified | M1 | |
| 60 | A1 | |
| Accept an embedded answer, eg \(\cos 60 = \dfrac{9}{18}\) with no further working | M1A1 |
| \(180 \div 3 = 60\) | M0A0 |
31 Use trigonometry to work out the value of \(x\). [2 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\cos 39 = \dfrac{x}{20}\) or \(20 \times \cos 39\) | M1 | oe eg \(\sin (90 - 39) = \dfrac{x}{20}\) or \(\sin 51 = \dfrac{x}{20}\) or \(20 \times \sin 51\) |
| 15.5(4…) | A1 | allow 16 with M1 seen |
| Alternative method 2 | ||
| \(20^2 - (20 \times \sin 39)^2\) | M1 | oe eg \(20^2 - (20 \times \cos 51)^2\) |
| 15.5(4…) | A1 | allow 16 with M1 seen |
| \(\cos = \dfrac{x}{20}\) unless recovered | M0 |
| \(20 \times 0.78\) | M1 |
| \(20 \times 0.78\) with an answer of 16 | M1A1 |
| \(20 \times 0.78\) with an answer of 15.6 | M1A0 |
| \(20 \times 0.77\) | M1 |
| \(20 \times 0.77\) with an answer of 16 | M1A1 |
| \(20 \times 0.77\) with an answer of 15.4 | M1A0 |
| cos (39 \(\times\) 20 unless recovered | M0 |
| Answer from scale drawing with no trigonometry | M0A0 |
19 Here is a cuboid.
DH = 8 cm
DB = 13 cm
Answer in degrees
Answer in degrees
| Answer | Mark | Comments |
|---|---|---|
| tan DBH \(= \dfrac{8}{13}\) | M1 | oe \(\tan^{-1}\dfrac{8}{13}\) |
| 31.6… | A1 |
| 31.6… in working, 32 on answer line – correct rounding | M1A1 |
| 31.6… in working, 31 on answer line – incorrect rounding | M1A0 |
| \(\tan\dfrac{8}{13}\) or \(\tan = \dfrac{8}{13}\) | M0A0 |
| Answer | Mark | Comments |
|---|---|---|
| 58.39… or 58.4 | B1ft | ft \(90 -\) their 31.6… |
Correct or follow through
30 Here is a right-angled triangle.
Not drawn accurately
Use trigonometry to work out the size of angle \(x\). [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\sin x = \dfrac{13}{16}\) or \(\sin^{-1} \dfrac{13}{16}\) | M1 | oe \(\sin x = 0.8125\) |
| 54(.3…) | A1 | |
| Alternative method 2 | ||
| \(\cos x = \dfrac{13}{16}\) or \(\cos^{-1} \dfrac{13}{16}\) and \(90 -\) their [35.6, 36] | M1 | oe |
| 54(.3…) | A1 | |
| Alternative method 3 | ||
| \(\cos x = \dfrac{\sqrt{16^2 - 13^2}}{16}\) or \(\tan x = \dfrac{13}{\sqrt{16^2 - 13^2}}\) | M1 | oe |
| 54(.3…) | A1 | |
| \(\sin = \dfrac{13}{16}\) or \(\sin \dfrac{13}{16}\) or \(\sin^{-1} = \dfrac{13}{16}\) unless recovered | M0 |
| Answer 54 from scale drawing with no trigonometry | M0A0 |
24 The length of a diagonal of a cuboid is 20 cm
The diagonal makes an angle of 24° with the base.
The area of the base is 150 cm2
Work out the volume of the cuboid. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sin 24 = \dfrac{h}{20}\) | M1 | oe \(\cos 66 = \dfrac{h}{20}\) \(\dfrac{20}{\sin 90} = \dfrac{h}{\sin 24}\) |
| \(20 \times \sin 24\) or 8.1… | M1dep | \(20 \times \cos 66\) \(\dfrac{20}{\sin 90} \times \sin 24\) |
| [1215, 1221] | A1 | with no incorrect working seen |
| \(150 \times 20 \times \sin 24\) | M1M1 |
19 Work out the area of the trapezium. [4 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| \(\tan 64 = \dfrac{h}{4}\) or \(\tan 26 = \dfrac{4}{h}\) or \(\dfrac{h}{\sin 64} = \dfrac{4}{\sin 26}\) | M1 | oe eg \(\tan 64 = \dfrac{h}{15 - 11}\) or \(\tan(90 - 64) = \dfrac{15 - 11}{h}\) or \(h^2 + 4^2 = \left(\dfrac{4}{\cos 64}\right)^2\) any letter |
| \(4 \tan 64\) or \(\dfrac{4}{\tan 26}\) or \(\dfrac{4}{\sin 26} \times \sin 64\) or 8.2… | M1dep | oe eg \(\sqrt{\left(\dfrac{4}{\cos 64}\right)^2 - 4^2}\) implies M2 may be seen on diagram |
| \(\dfrac{1}{2} \times (15 + 11) \times\) their 8.2… or \(\dfrac{1}{2} \times 4 \times\) their 8.2.. + \(11 \times\) their 8.2.. | M1dep | oe eg \(15 \times\) their 8.2… \(- \dfrac{1}{2} \times 4 \times\) their 8.2… dep on M2 |
| [106.6, 106.62] | A1 | accept 107 with working seen |
| 3rd M1 is for a total area and may be calculated as a trapezium or a rectangle + a triangle or a rectangle – a triangle or a triangle + a triangle | |
| 8.2… seen scores M2 even if not subsequently used | |
| Further work after 106.6 eg 106.6 + 16.4 | M1M1M0A0 |
29 Use trigonometry to work out the size of angle \(x\). [2 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| \(\cos x = \dfrac{9}{10}\) | M1 | oe eg \(\sin x = \dfrac{\sqrt{10^2 - 9^2}}{10}\) \(\tan x = \dfrac{\sqrt{10^2 - 9^2}}{9}\) |
| 25.8… or 26 | A1 |
| \(\cos = \dfrac{9}{10} \quad x = 25.8\) (recovered) | M1A1 |
| \(\cos = \dfrac{9}{10}\) | M0A0 |
25 \(ABC\) and \(ACD\) are triangles.
Not drawn accurately
Work out the size of angle \(x\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\tan 49 = \dfrac{AC}{16}\) | M1 | oe eg \(\tan (90 - 49) = \dfrac{16}{AC}\) or \(AC^2 + 16^2 = \left(\dfrac{16}{\cos 49}\right)^2\) |
| \(\tan 49 \times 16\) or [18.4, 18.41] | M1dep | oe eg \(\dfrac{16}{\tan (90 - 49)}\) or \(\sqrt{\left(\dfrac{16}{\cos 49}\right)^2 - 16^2}\) |
| \(\dfrac{\sin x}{\text{their } [18.4, 18.41]} = \dfrac{\sin 35}{20}\) or \(\dfrac{\text{their } [18.4, 18.41]}{\sin x} = \dfrac{20}{\sin 35}\) | M1dep | oe eg \(\dfrac{\sin x}{16 \tan 49} = \dfrac{\sin 35}{20}\) dep on 1st M1 |
| \(\sin x = \dfrac{\sin 35}{20} \times \text{their } [18.4, 18.41]\) | M1dep | oe eg \(\sin x = \dfrac{16 \tan 49 \sin 35}{20}\) or \(\sin^{-1}\left(\dfrac{\sin 35}{20} \times \text{their } [18.4, 18.41]\right)\) or \(\sin^{-1}\) [0.527, 0.528] dep on 1st and 3rd M1 |
| [31.8, 31.9] | A1 | allow 32 with full method seen |
| Answer [31.8, 31.9] possibly from scale drawing | 5 marks |
| Answer 32 possibly from scale drawing | Zero |
11 Use trigonometry to work out the size of angle \(x\). [2 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| \(\cos x = \dfrac{9}{10}\) | M1 | oe eg \(\sin x = \dfrac{\sqrt{10^2 - 9^2}}{10}\) \(\tan x = \dfrac{\sqrt{10^2 - 9^2}}{9}\) |
| 25.8… or 26 | A1 |
| \(\cos = \dfrac{9}{10} \quad x = 25.8\) (recovered) | M1A1 |
| \(\cos = \dfrac{9}{10}\) | M0A0 |
25 \(ABCDEF\) is a triangular prism which represents part of a hill.
\(ABCF\) is the horizontal rectangular base.
\(D\) is vertically above \(C\).
Work out the size of angle \(DAC\).
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\tan 6 = \dfrac{CD}{500}\) or \(500 \times \tan 6\) | M1 | oe any letter \(\dfrac{CD}{\sin 6} = \dfrac{500}{\sin 84}\) |
| [52.5, 52.6] or 53 | A1 | May be on diagram |
Check diagram for angle
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(500^2 + 400^2\) or \(250\,000 + 160\,000\) or \(410\,000\) | M1 | oe |
| \(\sqrt{\text{their } 410\,000}\) or \(\sqrt{500^2 + 400^2}\) or 640.(3…) | M1dep | \(AC\) |
| \(\tan x = \dfrac{[52.5, 52.6] \text{ or } 53}{\text{their } 640.(3..)}\) | M1dep | oe any letter |
| [4.6, 4.75] from correct working | A1 | accept 5 with correct working seen |
| Alternative method 2 | ||
| \(\dfrac{500}{\cos 6}\) or [502.7, 502.8] | M1 | oe \(BD\) |
| \(\sqrt{\left(\dfrac{500}{\cos 6}\right)^2 + 400^2}\) or [642.4, 642.5] | M1dep | \(AD\) |
| \(\sin x = \dfrac{[52.5, 52.6] \text{ or } 53}{\text{their } [642.4, 642.5]}\) | M1dep | oe any letter |
| [4.6, 4.75] from correct working | A1 | accept 5 with correct working seen |
| Alternative method 3 | ||
| \(500^2 + 400^2\) or \(250\,000 + 160\,000\) or \(410\,000\) or \(\dfrac{500}{\cos 6}\) or [502.7, 502.8] | M1 | oe \(BD\) |
| \(\sqrt{\text{their } 410\,000}\) or \(\sqrt{500^2 + 400^2}\) or 640.(3…) or \(\sqrt{\left(\dfrac{500}{\cos 6}\right)^2 + 400^2}\) or [642.4, 642.5] | M1dep | \(AC\) \(AD\) |
| \(\cos x = \dfrac{\text{their } 640.(3...)}{\text{their } [642.4, 642.5]}\) | M1dep | oe any letter |
| [4.6, 4.75] from correct working | A1 | accept 5 with correct working seen |
| Check diagram for lengths | |
| Beware \(\sin x = \dfrac{52.6}{640.(3...)}\) leads to [4.6, 4.75] | M1M1M0A0 |
19 A pentagon is made from a square and an isosceles triangle.
Not drawn accurately
Work out the perimeter of the pentagon. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Using one half of the isosceles triangle | ||
| (base angle =) 35 or (top angle =) 55 | B1 | may be on diagram |
| \(\cos(\text{their } 35) = \dfrac{6}{x}\) or \(\sin(\text{their } 55) = \dfrac{6}{x}\) or \(6^2 + (6 \tan(\text{their } 35))^2\) | M1 | oe eg \(\dfrac{\sin 90}{x} = \dfrac{\sin(\text{their } 55)}{6}\) any letter their 35 must be acute their 55 must be acute |
| \(\dfrac{6}{\cos(\text{their } 35)}\) or \(\dfrac{6}{\sin(\text{their } 55)}\) or \(\sqrt{6^2 + (6 \tan(\text{their } 35))^2}\) or 7.3(2…) | M1dep | oe |
| [50.6, 50.65] | A1ft | ft B0M2 with evaluation of \(36 + 2 \times\) their 7.3(2…) |
| Alternative method 2 Using the isosceles triangle | ||
| (base angle =) 35 or (top angle =) 110 | B1 | may be on diagram |
| \(\dfrac{x}{\sin(\text{their } 35)} = \dfrac{12}{\sin(\text{their } 110)}\) or \(12^2 = x^2 + x^2 - 2 \times x \times x \times \cos(\text{their } 110)\) or \(x^2 = x^2 + 12^2 - 2 \times x \times 12 \times \cos(\text{their } 35)\) | M1 | oe any letter their 35 must be acute their 110 cannot be 125 |
| \(\dfrac{12}{\sin(\text{their } 110)} \times \sin(\text{their } 35)\) or \(\sqrt{\dfrac{12^2}{2 - 2\cos(\text{their } 110)}}\) or \(\dfrac{12^2}{2 \times 12 \times \cos(\text{their } 35)}\) or 7.3(2…) | M1dep | oe |
| [50.6, 50.65] | A1ft | ft B0M2 with evaluation of \(36 + 2 \times\) their 7.3(2…) |
| Allow B1 even if the angle is not subsequently used | |
| Alt 2 Top angle 90 | M0M0A0 |
| Answer [50.6, 50.65] (possibly from scale drawing) | B1M1M1A1 |
29 Work out the size of angle \(x\). [2 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| \(\tan x = \dfrac{3}{7}\) or \(\tan^{-1} \dfrac{3}{7}\) or \(\sin x = \dfrac{3(\sin 90)}{\sqrt{3^2 + 7^2}}\) or \(\sin x = \dfrac{3(\sin 90)}{\sqrt{58}}\) or \(\cos x = \dfrac{7}{\sqrt{3^2 + 7^2}}\) or \(\cos x = \dfrac{7}{\sqrt{58}}\) or \(90 - \tan^{-1} \dfrac{7}{3}\) or 90 – [66.7, 66.81] or 90 – 67 | M1 | oe eg \(\cos x = \dfrac{7^2 + \left(\sqrt{7^2 + 3^2}\right)^2 - 3^2}{2 \times \sqrt{3^2 + 7^2} \times 7}\) Any letter |
| [23, 23.3] | A1 |
| \(\tan = \dfrac{3}{7}\) or \(\tan \dfrac{3}{7}\) or \(\tan^{-1} = \dfrac{3}{7}\) (unless recovered) | M0 |
| Answer [23, 23.3] (possibly coming from scale drawing) | M1A1 |
| If using sine rule must rearrange to \(\sin x =\) for M1 | |
| If using cosine rule must rearrange to \(\cos x =\) for M1 | |
| Allow [0.42, 0.43] for \(\dfrac{3}{7}\) | |
| Allow 2.33… for \(\dfrac{7}{3}\) | |
| Allow [7.6, 7.62] for \(\sqrt{3^2 + 7^2}\) |
27
VABCD is a square-based pyramid.The horizontal base ABCD has side length 10 cm and centre M.
Angle VMA is 90°
Angle VAM is 68°
Volume of pyramid \(= \dfrac{1}{3} \times\) area of base \(\times\) perpendicular height
Work out the volume of the pyramid. [6 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(10^2 + 10^2\) or 200 or \(5^2 + 5^2\) or 50 | M1 | oe |
| \(\sqrt{\text{their } 200}\) or \(10\sqrt{2}\) or [14, 14.2] or \(\sqrt{\text{their } 50}\) or \(5\sqrt{2}\) or [7, 7.1] | M1dep | oe |
| \(\tan 68 = \dfrac{h}{\text{their } 7.1}\) | M1dep | |
| their \(7.1 \times \tan 68\) or [17.3, 17.6] | M1dep | |
| \(\dfrac{1}{3} \times 10 \times 10 \times\) their [17.3, 17.6] | M1dep | |
| [576, 587] or 590 | A1 |
17 Work out the area of the parallelogram. [3 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\sin 72 = \dfrac{h}{12}\) or \(12 \sin 72\) or \(\cos (90 - 72) = \dfrac{h}{12}\) or \(12 \cos (90 - 72)\) or \(\dfrac{h}{\sin 72} = \dfrac{12}{\sin 90}\) or 11.4… | M1 | oe Any letter |
| \(16 \times\) their 11.4… | M1dep | |
| [182.4, 182.603] or 183 | A1 | |
| Alternative method 2 | ||
| \(h^2 + (12 \cos 72)^2 = 12^2\) or \(h^2 + (12 \sin (90 - 72))^2 = 12^2\) or \(\sqrt{12^2 - (12 \cos 72)^2}\) or \(\sqrt{12^2 - (12 \sin (90 - 72))^2}\) or 11.4… | M1 | oe Any letter |
| \(16 \times\) their 11.4… | M1dep | |
| [182.4, 182.603] or 183 | A1 | |
| Alternative method 3 | ||
| \(0.5 \times 16 \times 12 \times \sin 72\) or 91.3… | M1 | oe eg \(0.5 \times 16 \times 12 \times \sin 108\) |
| \(2 \times\) their 91.3… | M1dep | |
| [182.4, 182.603] or 183 | A1 | |
| \(2 \times 16 \times 12 \times \sin 72\) | M1M0A0 |
| \(\sin = \dfrac{h}{12}\) or \(\sin \theta = \dfrac{h}{12}\) (unless recovered) | M0 |
9 Work out the size of angle \(x\). [2 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| \(\tan x = \dfrac{3}{7}\) or \(\tan^{-1} \dfrac{3}{7}\) or \(\sin x = \dfrac{3(\sin 90)}{\sqrt{3^2 + 7^2}}\) or \(\sin x = \dfrac{3(\sin 90)}{\sqrt{58}}\) or \(\cos x = \dfrac{7}{\sqrt{3^2 + 7^2}}\) or \(\cos x = \dfrac{7}{\sqrt{58}}\) or \(90 - \tan^{-1} \dfrac{7}{3}\) or 90 − [66.7, 66.81] or 90 − 67 | M1 | oe eg \(\cos x = \dfrac{7^2 + \left(\sqrt{7^2 + 3^2}\right)^2 - 3^2}{2 \times \sqrt{3^2 + 7^2} \times 7}\) Any letter |
| [23, 23.3] | A1 |
| \(\tan = \dfrac{3}{7}\) or \(\tan \dfrac{3}{7}\) or \(\tan^{-1} = \dfrac{3}{7}\) (unless recovered) | M0 |
| Answer [23, 23.3] (possibly coming from scale drawing) | M1A1 |
| If using sine rule must rearrange to \(\sin x =\) for M1 | |
| If using cosine rule must rearrange to \(\cos x =\) for M1 | |
| Allow [0.42, 0.43] for \(\dfrac{3}{7}\) | |
| Allow 2.33… for \(\dfrac{7}{3}\) | |
| Allow [7.6, 7.62] for \(\sqrt{3^2 + 7^2}\) |
29 Use trigonometry to work out the length \(x\).
Not drawn accurately
[2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sin 72 = \dfrac{x}{8}\) or \(8 \times \sin 72\) or \(\cos(90 - 72) = \dfrac{x}{8}\) or \(8 \times \cos(90 - 72)\) or \(\dfrac{x}{\sin 72} = \dfrac{8}{\sin 90}\) or \(\dfrac{\sin 72}{x} = \dfrac{\sin 90}{8}\) | M1 | oe eg \(8\cos 72\) or 2.47… or 2.5 and \(\sqrt{8^2 - (8\cos 72)^2}\) |
| [7.6, 7.61] | A1 |
| If trigonometry and Pythagoras are used it must be a fully correct method that would lead to the correct value of \(x\) | |
| Accept \(\sin 72 \times 8\) | M1 |
| Accept opp or o for \(x\) \(\;\) eg \(\;\sin 72 = \dfrac{\text{opp}}{8}\) | M1 |
| \(\sin = \dfrac{x}{8}\) or \(\sin\theta = \dfrac{x}{8}\) (unless recovered) | M0 |
| Answer coming from scale drawing | M0A0 |
| Answer in range seen followed by 7 or 8 | M1A1 |
7 Use trigonometry to work out the length \(x\).
Not drawn accurately
[2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sin 72 = \dfrac{x}{8}\) or \(8 \times \sin 72\) or \(\cos (90 - 72) = \dfrac{x}{8}\) or \(8 \times \cos (90 - 72)\) or \(\dfrac{x}{\sin 72} = \dfrac{8}{\sin 90}\) or \(\dfrac{\sin 72}{x} = \dfrac{\sin 90}{8}\) | M1 | oe eg \(8 \cos 72\) or 2.47… or 2.5 and \(\sqrt{8^2 - (8\cos 72)^2}\) |
| [7.6, 7.61] | A1 |
| If trigonometry and Pythagoras are used it must be a fully correct method that would lead to the correct value of \(x\) | |
| Accept \(\sin 72 \times 8\) | M1 |
| Accept opp or o for \(x\) \(\;\) eg \(\sin 72 = \dfrac{\text{opp}}{8}\) | M1 |
| \(\sin = \dfrac{x}{8}\) or \(\sin \theta = \dfrac{x}{8}\) (unless recovered) | M0 |
| Answer coming from scale drawing | M0A0 |
| Answer in range seen followed by 7 or 8 | M1A1 |
24 Here is a right-angled triangle.
Not to scale
Use trigonometry to work out the value of \(x\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 57.9 to 58.03.. | 3 | M2 for \(\tan^{-1} \dfrac{7.2}{4.5}\) or M1 for \(\tan\ [x]\ [=]\ \dfrac{7.2}{4.5}\) or \(\tan\ [x]\ [=]\ 1.6\) oe | For M2 do not accept shift for \(\tan^{-1}\) Accept full methods using Pythagoras with sin or cos Don’t penalise If a different letter to \(x\) is used |
22 The diagram shows a square-based pyramid ABCDE.
O is the centre of the base.
The length of the base is 10.7 cm.
Angle EAO is 67°.
Calculate the length of AE.
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 19.3[4…] to 19.3[7…] or 19.4 with correct working | 6 | M2 for \(\left[\frac{1}{2} \times\right]\sqrt{10.7^2 + 10.7^2}\) or for \(\sqrt{5.35^2 + 5.35^2}\) or for \(\left[\frac{1}{2} \times\right] \frac{10.7}{\cos 45}\) or \(\left[\frac{1}{2} \times\right] \frac{10.7}{\sin 45}\) or for \(\frac{5.35}{\cos 45}\) or \(\frac{5.35}{\sin 45}\) or M1 for \(10.7^2 + 10.7^2\) or \(5.35^2 + 5.35^2\) or for \(\cos 45 = \frac{10.7}{\text{AC}}\) oe or \(\sin 45 = \frac{10.7}{\text{AC}}\) oe or for \(\cos 45 = \frac{5.35}{\text{AO}}\) oe or \(\sin 45 = \frac{5.35}{\text{AO}}\) oe | Correct working requires evidence of at least M1 AND M1 Accept intermediate values rot to at least 3sf. If M1A1 seen then award M2A1
| ||||||||||||||
| A1 for 7.56 to 7.57 or 7.6 AND | |||||||||||||||||
| M2 for \(\frac{\textit{their } 7.566\ldots}{\cos 67}\) or \(\frac{\textit{their } 7.566\ldots}{\sin 23}\) or \(\frac{\text{AE}}{\sin 90} = \frac{\textit{their } 7.566\ldots}{\sin 23}\) or M1 for \(\cos 67 = \frac{\textit{their } 7.566\ldots}{\text{AE}}\) oe or \(\sin 23 = \frac{\textit{their } 7.566\ldots}{\text{AE}}\) oe or \(\frac{\sin 90}{\text{AE}} = \frac{\sin 23}{\textit{their } 7.566}\) oe | their 7.566…must follow from the award of the first M2 | ||||||||||||||||
| If 0, 1 or 2 scored, instead award SC3 for answer 19.3 to 19.4 with no or insufficient working If 0 or M1 scored, instead award SC2 for 7.56 to 7.57 or 7.6 with no or insufficient working If 0 scored instead award SC1 for 15.1[3…] or 57.2… with no or insufficient working | Mark less efficient methods by considering how many steps remain to reach the correct answer and match against the most comparable stage: eg Assuming AO has been found, then using trig to find EO is unnecessary and inefficient, so no further marks are awarded for just finding EO and stopping. However, if EO is then used in a correct Pythagoras expression with AO or a correct trig expression with 67° that lead directly or indirectly to finding AE, then M2 or M1 may be awarded. | ||||||||||||||||
23 The diagram shows two right-angled triangles, ABC and BDC, joined at BC.
Not to scale
Work out angle BCA.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 60 [\(^\circ\)] with correct working | 5 | Correct working requires evidence of at least M2A1M1 | |
| M2 for \(\sqrt{12^2 + 5^2}\) oe or M1 for \(5^2 + 12^2\) oe A1 for [BC = ] 13 | Condone stating 5, 12, 13 or \(5^2 + 12^2 = 13^2\) for M2 After M1, [BC =] 13 implies M2A1 [BC =] 13 might be on the diagram Do not penalise if Pythagoras is not subsequently used in work leading to their answer. | ||
| M1 for \(\dfrac{\textit{their}\ 13}{26} = \cos\) […] oe | |||
| If 0 or 1 scored, instead award SC2 for final answer of 60 nfww If 0 scored, instead award SC1 for [BC =] 13 | Note that: \(5 \times 12 = 60\) is SC0 | ||
21 The diagram shows a cuboid ABCDEFGH.
Not to scale
FB = 15 cm, GF = 7 cm and HG = 24 cm.
Calculate the angle BHF.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 30.9 to 31 with correct working | 5 | M2 for \(\sqrt{(24^2 + 7^2)}\) or M1 for \([\ldots]^2 = 24^2 + 7^2\) and M2 for \(\tan^{-1} \frac{15}{\textit{their } 25}\) oe or M1 for \(\tan(\text{BHF}) = \frac{15}{\textit{their } 25}\) oe | Correct working requires evidence of at least M1M1 Alternative method: M2 for \(\sqrt{(24^2 + 7^2 + 15^2)}\) or M1 for \([\ldots]^2 = 24^2 + 7^2 + 15^2\) and M2 for \(\sin^{-1} \frac{15}{\textit{their } \sqrt{850}}\) oe or M1 for \(\sin(\text{BHF}) = \frac{15}{\textit{their } \sqrt{850}}\) oe Other methods: M2 for a correct expression that would give correct HF or HB or M1 for “one-step away” eg \(\sin\left(\tan^{-1}\left(\frac{7}{24}\right)\right) = \frac{7}{\text{HF}}\) and M2 for correct trig inverse expression with angle BHF or M1 for correct trig expression with angle BHF |
| If 0 or 1 scored, instead award SC2 for answer 30.9 to 31 with no or insufficient working If 0 scored, instead award SC1 for [HF =] 25 or for [HB =] \(\sqrt{850}\) oe or 29.1 to 29.2 | May be on diagram | ||
16 AC is a diagonal of the quadrilateral ABCD.
Not to scale
AD = 72 m.
Angle ABC = 108°, angle BCA = 38° and angle BAC = 34°.
Angle ACD = 90°, angle CDA = 65° and angle CAD = 25°.
Find the area of ABCD.
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 1763 or 1763.4 to 1763.5 with correct working | 6 | “Correct working” requires evidence of at least M2 or M1 M1 condone answers in range 1750 – 1780 with full correct working | |
| M2 for \(\frac{1}{2} \times 72 \times\) their 65.25… × sin 25 or M1 for [AC=] 72cos 25 or 72sin 65 oe | M2 implied by 992.79… or 992.8 or maybe by 985 – 999[…] nfww or M1 implied by [AC=] 65.2[5…] or 65.3 | ||
| M3 for \(\frac{1}{2} \times\) their 65.254 × their 42.24… × sin34 oe or M2 for [AB=] \(\frac{\textit{their } 65.25\ldots \times \sin 38}{\sin 108}\) oe or M1 for \(\frac{\text{AB}}{\sin 38} = \frac{\textit{their } 65.25\ldots}{\sin 108}\) oe | M3 implied by 770.69… or 770.7 or maybe by 760 – 783 nfww M2 implied by [AB=] 42.2[4…] | ||
| If 0, 1 or 2 scored award SC3 for answer 1763 or 1763.4 to 1763.5 with no or insufficient working If 0 or 1 scored award SC2 for 992.79… or 992.8 or 770.69… or 770.7 with no or insufficient working | accept any correct method e.g. M2 for area of triangle ACD = ½ × 72sin25 × 72cos25 oe or M1 for [CD=]72sin25 implied by 30.4… | ||
Area of ABC = 770.687099… and area of ACD = 992.793598… total area = 1763.480697…
Note : condone answers in range 1745 – 1780, area ACD in range 985 – 998 and area ABC in range 760 – 783
22 The diagram shows a regular pentagon made using ten congruent right-angled triangles.
The length of one side of the pentagon is 12 cm.
Not to scale
The pyramid has volume 450 cm\(^3\).
The perpendicular height of the pyramid is \(h\) cm.
Calculate the value of \(h\).
[The volume of a pyramid is \(\dfrac{1}{3} \times\) area of base \(\times\) perpendicular height.] [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Allocate marks similarly for other methods such as five triangles using an angle of 72. If in doubt consult TL There must be evidence of angle or trig work to score any marks e.g. working back from 247.75 to \(h = 8.258\)… is likely to score 0 or B1 | |||
| [angle =] 36 or 54 | B1 | in correct place if only shown on diagram | Do not award 36 or 54 if calculated as an area |
| [\(h =\)] \(\dfrac{6}{\tan 36}\) or \(6 \times \tan 54\) or \(\dfrac{6 \sin 54}{\sin 36}\) may be implied by 8.258 to 8.259 or 8.26 following M1 but not from area | M2 | M1 for \(\tan 36 = \frac{6}{h}\) or \(\tan 54 = \frac{h}{6}\) or \(\frac{6}{\sin 36} = \frac{h}{\sin 54}\) or \(\frac{\sin 36}{6} = \frac{\sin 54}{h}\) | Accept other notation for ‘h’ |
| \(10 \times \frac{1}{2} \times 6 \times\) their \(h\) oe | M2dep | M1 for \(\frac{1}{2} \times 6 \times\) their \(h\) oe may be implied by 24.774 to 24.775 | Their h dep on previous M2 or M1 Accept correct use of \(\frac{1}{2}ab\sin C\) |
| 247.748 to 247.749 | A1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 5.45 or 5.449 to 5.450 nfww | 3 | M2 for \(h = \frac{450 \times 3}{247.75}\) oe or M1 for \(\frac{1}{3}h \times 247.75 = 450\) oe | 247.75 may be their more accurate 247.748 to 247.749 or 247.7 247.8 or 247.74 from (a) Use of incorrect formula is not MR |
5 The diagram shows a regular pentagon made using ten congruent right-angled triangles.
The length of one side of the pentagon is 12 cm.
Not to scale
The pyramid has volume \(450\,\text{cm}^3\).
The perpendicular height of the pyramid is \(h\) cm.
Calculate the value of \(h\).
[The volume of a pyramid is \(\frac{1}{3} \times \text{area of base} \times \text{perpendicular height}\).] [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Allocate marks similarly for other methods such as five triangles using an angle of 72. If in doubt, consult TL. There must be evidence of angle or trig work to score any marks e.g. working back from 247.75 to \(h = 8.258\)…is likely to score 0 or B1 | |||
| [angle =] 36 or 54 | B1 | in correct place if only shown on diagram | Do not award 36 or 54 if calculated as an area |
| [\(h =\)] \(\frac{6}{\tan 36}\) or 6 × tan54 or \(\frac{6\sin 54}{\sin 36}\) may be implied by 8.258 to 8.259 or 8.26 following M1 but not from area | M2 | M1 for \(\tan 36 = \frac{6}{h}\) or \(\tan 54 = \frac{h}{6}\) or \(\frac{6}{\sin 36} = \frac{h}{\sin 54}\) or \(\frac{\sin 36}{6} = \frac{\sin 54}{h}\) | Accept other notation used for ‘\(h\)’ |
| \(10 \times \frac{1}{2} \times 6 \times \textit{their } h\) oe | M2dep | M1 for \(\frac{1}{2} \times 6 \times \textit{their } h\) oe may be implied by 24.774 to 24.775 | Their \(h\) dep on previous M2 or M1 Accept correct use of \(\frac{1}{2}ab\sin C\) |
| 247.748 to 247.749 | A1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 5.45 or 5.449 to 5.450 nfww | 3 | M2 for \(h = \frac{450 \times 3}{247.75}\) oe or M1 for \(\frac{1}{3}h \times 247.75 = 450\) oe | 247.75 may be their more accurate 247.748 to 247.749 or 247.7, 247.8 or 247.74 from (a) Use of incorrect formula is not MR |
21
They tell the teacher that \(\sin B = \dfrac{5}{4}\).
Explain why this student must be wrong. [1]
Not to scale
Work out the value of \(x\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Eg sin must be \(\leqslant 1\) their opp is \(\gt\) than their hyp | 1 | See appendix | |
Exemplar responses for Q21a
| Response | Mark |
|---|---|
| The fraction cannot be bigger than 1 | 1 |
| The numerator cannot be bigger than the denominator | 1 |
| Not using degrees with sin and putting B in the wrong place | 0 |
| The student must be wrong because the question said so | 0 |
| That is upside down \(\frac{4}{5}\) | 0 |
| It’s the wrong calculation | 0 |
| The fraction cannot be bigger than 1 otherwise it’s a reflex angle in a triangle which is not possible | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 53.1[3…] | 3 | M2 for \(\cos^{-1} \frac{24}{40}\) or M1 for \(\cos\) [\(x\)] [=] \(\frac{24}{40}\) or 0.6 | Accept 53 after M1 or M2 scored Accept full methods using Pythagoras or sin or tan |
25 Taylor designs a logo using isosceles triangles joined at a central point, P.
This is the start of Taylor’s design.
Not to scale
The completed design will have rotational symmetry, order 60 about point P.
Each triangle has base, \(b\), and height, \(h\), measured in mm.
Not to scale
Calculate \(h\) when \(b = 40\) mm.
Give your answer correct to 1 decimal place. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.1[0…] nfww | 4 | M1 for \(\dfrac{360}{60}\) oe soi by 6 AND Method 1 using tan: M2 for [\(h\) = ] 20 tan(their 6) oe or M1 for correct use of tan(their 6) oe or Method 2 using sine rule: M2 for [\(h\) = ] \(\dfrac{20\sin(\textit{their}\ 6)}{\sin(90 - \textit{their}\ 6)}\) or M1 for \(\dfrac{\sin(\textit{their}\ 6)}{h} = \dfrac{\sin(90 - \textit{their}\ 6)}{20}\) oe or Method 3 using cos and Pythagoras: M2 for \(\sqrt{\left(\dfrac{20}{\cos(\textit{their}\ 6)}\right)^2 - 20^2}\) or M1 for \(\left(\dfrac{20}{\cos(\textit{their}\ 6)}\right)^2 - 20^2\) | May be on diagram In all methods, if their angle is not 6 then method must be seen, not implied by interim answers unless stated otherwise Accept any acute angle used for their 6 eg [\(h\) = ] \(\dfrac{20}{\tan(90 - \textit{their}\ 6)}\) eg tan(their 6) = \(\dfrac{h}{20}\) NBs \(\dfrac{\text{approx. circumference}}{60} = \dfrac{40\pi}{60} = 2.1\) scores 0 20sin6 = 2.1 scores M1 for 6 Solution from scale drawing scores a maximum of M1 if 6 seen |
22 The diagram shows a right-angled triangle.
Not to scale
Work out the value of \(a\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7 | 3 | B1 for \(\sin 30 = \dfrac{1}{2}\) oe M1 for \(\sin 30 = \dfrac{a}{14}\) or better | Accept any letter or correct identification for \(a\) e.g. \(x\) or opposite M1 implied by 14 x sin 30 oe Alternative Method: B1 for \(\cos 60 = \dfrac{1}{2}\) oe M1 for \(\cos 60 = \dfrac{a}{14}\) or better Accept equivalent Sine Rule application |
13 The diagram shows a triangle, ABC, with perpendicular height BD.
Not to scale
BC = 12 cm, angle BCD = 30° and angle BAD = 45°.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 6 | 3 | M1 for sin 30 = \(\frac{BD}{12}\) oe B1 for sin 30 = 0.5 oe soi | Accept e.g. sin 30 = \(\frac{\sqrt{1}}{2}\) Do not allow B1 if contradicted |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(6\sqrt{2}\) or \(\frac{12}{\sqrt{2}}\) cao | 3 | B2 for \(\sqrt{72}\) or better OR M1 for \(\frac{their(a)}{\sin 45}\) [= \(AB\)] oe or 2(their (a))\(^2\) oe or \(\frac{12 \times \sin 30}{\sin 45}\) [= \(AB\)] B1 for sin 45 = \(\frac{\sqrt{2}}{2}\) or \(\frac{1}{\sqrt{2}}\) oe soi | For M1, accept \(\frac{their(a)}{\cos 45}\) [= \(AB\)] oe For M1 , accept e.g. \(\frac{their(a) \times \sin 90}{\sin 45}\) [= \(AB\)] M1 implied by e.g. 72 For B1 accept cos 45 = \(\frac{\sqrt{2}}{2}\) or \(\frac{1}{\sqrt{2}}\) oe Do not allow B1 if contradicted |
8 Taylor designs a logo using isosceles triangles joined at a central point, P.
This is the start of Taylor’s design.
Not to scale
The completed design will have rotational symmetry, order 60 about point P.
Each triangle has base, \(b\), and height, \(h\), measured in mm.
Not to scale
Calculate \(h\) when \(b = 40\) mm.
Give your answer correct to 1 decimal place. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.1[0…] nfww | 4 | M1 for \(\frac{360}{60}\) oe soi by 6 AND Method 1 using tan: M2 for [\(h\) = ] 20 tan(their 6) oe or M1 for correct use of tan(their 6) oe or Method 2 using sine rule: M2 for [\(h\) = ] \(\frac{20 \sin(their\ 6)}{\sin(90 - their\ 6)}\) or M1 for \(\frac{\sin(their\ 6)}{h} = \frac{\sin(90 - their\ 6)}{20}\) oe or Method 3 using cos and Pythagoras: M2 for \(\sqrt{\left(\frac{20}{\cos(their\ 6)}\right)^2 - 20^2}\) or M1 for \(\left(\frac{20}{\cos(their\ 6)}\right)^2 - 20^2\) | May be on diagram In all methods, if their angle is not 6 then method must be seen, not implied by interim answers unless stated otherwise Accept any acute angle used for their 6 eg [\(h\) = ] \(\frac{20}{\tan(90 - their\ 6)}\) eg tan(their 6) = \(\frac{h}{20}\) NBs \(\frac{\text{approx. circumference}}{60} = \frac{40\pi}{60}\) = 2.1 scores 0 20sin6 = 2.1 scores M1 for 6 Solution from scale drawing scores a maximum of M1 if 6 seen |
25 The diagram shows an isosceles triangle, ABC.
Not to scale
AB = AC = 19 cm.
Angle BAC = 54°.
Calculate the length of BC.
You must show your working.
.................... cm [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 17 or 17.2 to 17.3 with correct working | 5 | M4 for \((\sin 27 \times 19) \times 2\) or \((\cos 63 \times 19) \times 2\) or M3 for \(\sin 27 \times 19\) or \(\cos 63 \times 19\) or M2 for \(\sin 27 = \dfrac{x}{19}\) oe or \(\cos 63 = \dfrac{x}{19}\) oe or B1 for 27 as angle If 0 or 1 scored, instead award SC2 for 17.2 to 17.3 with no or insufficient working If 0 scored, instead award SC1 for 8.6 or 8.62 to 8.63 or 17 with no or insufficient working | “Correct working” requires evidence of at least M2 May be seen in the correct place on the diagram Accept fully correct alternative methods |
11 The diagram shows two right-angled triangles that are joined together.
All measurements are given accurate to 2 significant figures.
Not to scale
Work out the value of \(x\).
Give your answer correct to an appropriate degree of accuracy.
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 9.6 with correct working | 6 | M5 for \(\sqrt{8.3^2 + (5.4\tan 42)^2}\) OR M2 for 5.4 × tan 42 oe or M1 for tan 42 = \(\frac{opp}{5.4}\) oe A1 for 4.8[6…] or 4.9 and M2 for \(\sqrt{8.3^2 + (their\ 4.8[6\ldots])^2}\) or M1 for \(8.3^2 + (their\ 4.8[6\ldots])^2\) If 0, 1 or 2 scored, instead award SC3 for answer 9.6 with no or insufficient working If 0 or 1 scored, instead award SC2 for answer 9.619… or 9.62 with no or insufficient working If 0 scored, instead award SC1 for 4.8[6…] or 4.9 with no or insufficient working | “Correct working” requires evidence of at least M1 M1 or M2 and also accept 2sf numbers 4.8[6…] or 4.9 can imply M2 if M1 given Alternative method (sine rule): M2 for \(\frac{5.4 \times \sin 42}{\sin(their\ 48)}\) oe or M1 for \(\frac{x}{\sin 42} = \frac{5.4}{\sin(their\ 48)}\) oe 9.619… or 9.6[2] can imply M2 if M1 awarded Allow full and correct equivalent trig. methods to score M2 in both triangles e.g. using the other angle in the first triangle, i.e. their 48 |
19
[1]
Not to scale
Work out the value of \(x\). [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{1}{2}\) | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 12 | 2FT | B2FT for final answer \(12\sqrt{3}\), 8, \(8\sqrt{3}\) or 6 if consistent with their (a) or M1 for \(24 \times \dfrac{1}{2}\) or \(24 \times\) their (a) | For 1 or 2 marks, isw attempts to simplify the surds May be implied by their answer Ignore units |
23 Here is a right-angled triangle.
Not to scale
Work out the value of \(h\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 16 nfww | 3 | M2 for [\(h =\)] \(\dfrac{8}{\cos 60}\) or M1 for \(\cos 60 = \dfrac{8}{h}\) | Accept alternative methods e.g. M2 for [\(h =\)] \(\dfrac{8}{\sin 30}\) or M1 for \(\sin 30 = \dfrac{8}{h}\) |
19 ABC is an isosceles triangle.
The sides of the triangle ABC are all tangents to a circle of radius 6 cm, centre O.
Not to scale
Angle BAC = 70° and BA = BC.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| complete correct argument e.g. | accept any correct method not using 17.54 | ||
| angle ABC = 40° | B1 | could be on diagram and also accept ABO = 20°, BO’T’ = 70° | |
| [BO = ] e.g. \(\dfrac{6}{\sin 20}\) | M2 | M1 for e.g \(\sin 20 = \dfrac{6}{[BO]}\) | i.e BO as subject for M2 and condone sine rule with sin 90° for M2 |
| 17.542… or 17.543 | A1dep | dep. on at least M1 | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 202 or 201.5 to 201.8 with correct working | 5 | Accept any correct method e.g. M1 for [height=] 17.54 + 6 or 23.54… M2 for [half base =] \(\dfrac{6}{\tan 35}\) or \(\dfrac{23.54}{\tan 70}\) or M1 for \(\tan 35 = \dfrac{6}{\textit{half base}}\) or \(\tan 70 = \dfrac{23.54}{\textit{half base}}\) | “Correct working” requires evidence of at least M2 or M1M1 Condone use of 8.6 leading to an answer of 202.4… M2 implied by e.g. 8.56 to 8.57 or 17.1 to 17.2 |
| M1 for \(\frac{1}{2}\) × their base × their height oe If 0 scored SC2 for 202 or 201.6 to 201.8 with no working or SC1 for 8.56 to 8.57 or 17.1 to 17.2 with no working | e.g. \(\frac{1}{2}\) × (2 × 8.568…) × 23.54… | ||
11 The diagram shows a quadrilateral ABCD.
AD = 10 cm, angle BAD = 30° and angle ABD = angle BDC = 90°.
Not to scale
The ratio of length BD to length DC is 1 : 2.4 .
Work out length BC.
You must show your working. [7]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 13 with correct working | 7 | M1 for \(\dfrac{BD}{10} = \sin 30\) oe B1 for sin 30 = 0.5 soi A1 for [BD =] 5 M1 for their BD × 2.4 oe A1 for CD = 12 | “Correct working” requires evidence of at least M1 or B1 and M1M1 or alternate convincing approach Answer 12 gets A0 unless CD = 12 shown in working or on diagram |
| M1 for (their BD)² + (their CD)² [ = BC²] oe If 0 scored SC3 for answer 13 with no working or SC2 for CD = 12 with no working or SC1 for BD = 5 with no working | SCs may be seen on the diagram | ||
18 Here is a right-angled triangle.
Not to scale
Use trigonometry to work out the value of \(x\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 5.39[6...] or 5.4[0] | 3 | M2 for \(8 \times \tan 34\) or any complete correct method or M1 for \(\tan 34 = \dfrac{x}{8}\) | e.g. \(\dfrac{8}{\tan(90 - 34)}\) e.g. \(\tan(90 - 34) = \dfrac{8}{x}\) See appendix |
Exemplar responses for Q18
| Response | Mark |
|---|---|
| For longest method M1 is for step before explicit calculation for answer \(\cos 34 = \dfrac{8}{y}\) so \(y = \dfrac{8}{\cos 34}\) so \(x^2 = \left(\dfrac{8}{\cos 34}\right)^2 - 8^2\) (corrected from the printed mark scheme, which shows \(\left(\dfrac{y}{\cos 34}\right)^2 - 8^2\)) | M1 |
| For longest method M2 is for explicit calculation for answer \(\cos 34 = \dfrac{8}{y}\) so \(y = \dfrac{8}{\cos 34}\) so \(x = \sqrt{\left(\dfrac{8}{\cos 34}\right)^2 - 8^2}\) | M2 |
Allow e.g. for M1 | M1 |
16 The diagram shows a circle, centre O.
Points A, B and C lie on the circumference of the circle.
Line AOB is a diameter.
Line DAE is a tangent to the circle.
Angle CAE = 32°.
Not to scale
Calculate length BC. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [angle in a] semi-circle oe | 1 | Accept other reasoning if fully justified | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 13.5 to 13.6 | 4 | B1 for angle BAC = 58° or angle ABC = 32° M2 for 16sin(their 58) or 16cos(their 32) or M1 for \(\sin(\textit{their}\ 58) = \frac{BC}{\textit{their}\ 16}\) or \(\cos(\textit{their}\ 32) = \frac{BC}{\textit{their}\ 16}\) or better If 0 or B1 scored then instead award SC2 for 6.7 to 6.8 as final answer Grads or rads: If 0, 1 or 2 scored then instead award SC3 for 15.8[8…] to 15.9 or 12.6[4…] as final answer or If 0 scored award SC1 for 7.9[4…] or 6.3[2…] | May be seen on diagram or implied by use of sin58 or cos32 Only award M marks if their angle and trig ratio are consistent ie do not accept 16sin32 unless angle BAC previously seen as 32. |
19
\(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\)
Not to scale
Work out the length of BC. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\frac{1}{\sqrt{2}}\) or \(\frac{\sqrt{2}}{2}\) final answer | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 10 nfww | 6 | B3 for BD = \(10\sqrt{2}\) oe or M2 for \(10\sqrt{6} \times \tan 30\) oe or M1 for \(\frac{BD}{10\sqrt{6}} = \tan 30\) oe AND M2 for BC = \(\sqrt{\frac{\textit{their}\ BD^2}{2}}\) oe or their BD × their sin 45 oe or M1 for BC2 + CD2 = (their BD)2 or for \(\frac{BC}{\textit{their}\ BD} = \textit{their}\ \sin 45\) | Allow use of other variables for BC and CD (possibly different) |
13 Here is a right-angled triangle.
Not to scale
Show that angle \(x\) is \(35^\circ\), correct to the nearest degree. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Shows correct working leading to 34.9[9...] seen [rounding to 35] | 3 | M2 for \(\tan^{-1} \dfrac{14}{20}\) or M1 for tan [=] \(\dfrac{14}{20}\) or tan[=] 0.7 or tan[x] [=] \(\dfrac{14}{20}\) or 0.7 | If using Pythagoras , sin or cos, must have full method. Accept change of variable |
23 The diagram shows a regular hexagon made from six equilateral triangles.
Each side is 10 cm.
The angle ACB is a right angle.
Not to scale
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 60 or 30 seen as angle | B1 | May be correctly marked on diagram | Reverse method using 8.66... scores 0 |
| \(10 \times \sin 60\) or \(10 \times \cos 30\) | M2 | M1 for \(\sin 60 = \dfrac{AC}{10}\) oe or \(\cos 30 = \dfrac{AC}{10}\) | |
| 8.660[...] | A1 dep | Dep on at least M1 | |
| Alternative method by Pythagoras | |||
| 5 seen as side | B1 | May be correctly marked on diagram | |
| \(\sqrt{10^2 - 5^2}\) | M2 | or M1 for \(10^2 - 5^2\) | \(10^2\) may be 100 and \(5^2\) may be 25 |
| 8.660[...] | A1 dep | Dep on at least M1 | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) \(\dfrac{1}{2} \times \dfrac{1}{2} \times 10 \times 8.66[0..]\) oe | M1 | Reverse method using 21.7 scores 0 May be in stages | |
| 21.65[...] | A1 | ||
| (ii) 260 | 2 | M1 for \(12 \times 21.7\) or B1 for 259.8 to 260.4 | Award M1 for alternative complete methods |
5 ABC is a right-angled triangle.
AB = 20 cm and BC = 37 cm.
Not to scale
Calculate angle BAC. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Answer which rounds to 61.6 nfww | 3 | M2 for \(\tan^{-1}\left(\dfrac{37}{20}\right)\) oe or M1 for tan[x =] \(\dfrac{37}{20}\) oe If M0 scored then SC1 for answers 28.4, 28 or angles that round to 28.4 if correct working seen. | Condone answer of 62 only if correct working seen Answers of 68.5 or 68.4(5..) [grads] or 1.08 or 1.07(5..) [rads] imply M2 Alternative method After correct method for Pythagoras soi by 42.0 to 42.1 M2 for \(\sin^{-1}\left(\dfrac{37}{\textit{their}\sqrt{20^2 + 37^2}}\right)\) or \(\cos^{-1}\left(\dfrac{20}{\textit{their}\sqrt{20^2 + 37^2}}\right)\) or M1 for sin[x =] \(\dfrac{37}{\textit{their}\sqrt{20^2 + 37^2}}\) or cos[x =] \(\dfrac{20}{\textit{their}\sqrt{20^2 + 37^2}}\) or M0 for just Pythagoras reaching AC = 42.0 to 42.1 Do not condone answer of 62 following an interim answer seen that does not round to 61.6 0 for scale drawing |
21 Here is a right-angled triangle.
Use trigonometry to work out the value of \(x\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 3.488 to 3.489 or 3.49 or 3.5 | 3 | M2 for 10.2 \(\times\) sin20 or any complete correct method or M1 for \(\sin 20 = \dfrac{x}{10.2}\) | Cos70 \(\times\) 10.2 or 10.2 \(\times\) cos 20 and \(\sqrt{10.2^2 - (10.2 \times \cos 20)^2}\) Allow 10.2 \(\times\) cos 20 with attempt at Pythagoras for M1 |
13 In the diagram, ABC is a right-angled triangle.
P is a point on AB.
BC = 40 m, AP = 20 m and angle ABC = 30°.
Not to scale
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| AC = 40sin30 | M2 | M1 for \(\dfrac{\text{AC}}{40} = \sin 30\) oe | |
| 20 and evidence that sin 30 = 0.5 | A1 | If 0 scored, B1 for sin 30 = 0.5 oe | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(20(\sqrt{3} - 1)\) | 5 | B4 for \(20\sqrt{3} - 20\) or \(\sqrt{1200} - 20\) or B3 for \(\sqrt{1200}\) or \(\dfrac{40\sqrt{3}}{2}\) or M2 for 40cos30 oe or M1 for \(\cos 30 = \dfrac{\text{AB}}{40}\) oe | Other methods are possible e.g. Pythag e.g. M2 for [AB =] \(\sqrt{40^2 - 20^2}\) e.g. M1 for AB² + 20² = 40² |
11 The diagram shows two right-angled triangles ABD and BCD, sharing a common side BD.
AD = 10 cm, BC = 12 cm and angle DBC = 60°.
Not to scale
Work out the length of AB. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 8 nfww | 6 | nfww - must check method before giving 6 marks must not come from wrong working | |
| B1 for cos 60 = 0.5 oe soi M1 for \(\dfrac{\text{BD}}{12} = \cos 60\) oe A1 for [BD =] 6 AND M2 for \(\sqrt{10^2 - \textit{their}\text{ BD}^2}\) or M1 for their \(\text{BD}^2 + \text{AB}^2 = 10^2\) oe or for \(10^2 -\) their \(\text{BD}^2\) | e.g. sin 30 = 0.5 e.g. \(\dfrac{\text{BD}}{12} = \sin 30\) For M2 or M1 0 < their BD < 10 and BD must be identified first on diagram or in working | ||
8 A, B, C and D are points on the circumference of a circle, centre O.
Not to scale
Angle CAD = 28° and CD = 6.4 cm.
BD is a diameter of the circle.
Calculate the area of the circle. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 145.2 to 146.2 | 5 | B1 for angle CBD = 28 soi or for angle BCD = 90 soi and M2 for \(\dfrac{6.4}{\sin 28}\) oe or 13.6[3…] nfww or M1 for \(\sin [28] = \dfrac{6.4}{[..]}\) oe and M1 for \(\pi \times (\textit{their}\text{ radius})^2\) | B1 implied by e.g. 28 or 62 correctly used in trigonometry or 28, 62 or 90 (or symbol) marked in the correct place in the diagram 13.6… can imply B1 however if it is marked on the wrong side, e.g. on AC, then it scores 0 marks |
19 In the diagram below, ABCD is a trapezium.
Not to scale
Length AE is 37.5 cm.
DE = CF
Find the value of angle \(x\). [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 38.7 | 6 | B3 for 50 for DE or CF nfww Or M1 for \(62.5^2 - 37.5^2\) M1 for \(\sqrt{62.5^2 \pm 37.5^2}\) And B3 FT for \(\sin^{-1}\left(\dfrac{\textit{their } 50}{80}\right)\) correctly evaluated or M2 FT for \(\sin^{-1}\left(\dfrac{\textit{their } 50}{80}\right)\) or M1 FT for \(\sin\) [\(x\)] = \(\dfrac{\textit{their } 50}{80}\) | Allow 39 May be in correct place on diagram 2500 implies M1 |
19 The angles in a triangle are in the ratio 1 : 2 : 3.
Calculate the length of the shortest side in the triangle. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 180 \(\div\) (1 + 2 + 3) \(\times\) 3 [= 90] | 2 | M1 for 180 \(\div\) (1 + 2 + 3) If 0 scored, SC1 for angles 30, 60, 90 | Condone 6 for 1 + 2 + 3 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7.5 | 4 | B1 for sin 30° or cos 60° = \(\frac{1}{2}\) soi M2 for 15 sin 30 oe or M1 for \(x/15\) = sin 30 oe | |
8 The angles in a triangle are in the ratio 1 : 2 : 3.
Calculate the length of the shortest side in the triangle.
........................................ cm [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 180 ÷ (1 + 2 + 3) × 3 [= 90] | 2 | M1 for 180 ÷ (1 + 2 + 3) If 0 scored, SC1 for angles 30, 60, 90 | Condone 6 for 1 + 2 + 3 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7.5 | 4 | B1 for sin 30° or cos 60° = ½ soi M2 for 15 sin 30 oe or M1 for \(x/15\) = sin 30 oe | |
22 A, B, C and D are four towns.
B is 25 kilometres due East of A.
C is 25 kilometres due North of A.
D is 45 kilometres due South of A.
Not to scale
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 135 | 2 | B1 for angle 45 | e.g 45 marked at ACB or ABC or 180 – 45 or 90 + 45 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 209 to 209.1 | 4 | M2 for \(\tan^{-1}(45 \div 25)\) or \(\tan^{-1}(25 \div 45)\) soi by 61, 60.94 to 60.95 or 29[.1], 29.05... or M1 for \(\tan\) [=] \(45 \div 25\) or \(\tan\) [=] \(25 \div 45\) AND M1 for 270 – their angleABD or 180 + their angleADB | Accept longer methods but they must get to the equivalent point to gain credit e.g. if they find the hypotenuse, they score M0 until they start to use sin or cos. Can be implied by their answer |
7 A, B, C and D are four towns.
B is 25 kilometres due East of A.
C is 25 kilometres due North of A.
D is 45 kilometres due South of A.
Not to scale
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 135 | 2 | B1 for angle 45 | e.g. 45 marked at ACB or ABC, 180 – 45, 90 + 45 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 209 to 209.1 | 4 | M2 for \(\tan^{-1}(45 \div 25)\) or \(\tan^{-1}(25 \div 45)\) soi by 61, 60.94 to 60.95 or 29[.1], 29.05... or M1 for tan [=] 45÷25 or tan [=] 25÷45 | Accept longer methods but they must get to the equivalent point to gain credit e.g. if they find the hypotenuse, they score M0 until they start to use sin or cos. Can be implied by their answer |
| AND | |||
| M1 for 270 – their angleABD or 180 + their angleADB | |||
No questions match these filters.