Find the size of angle \(ABC\). Give your answer correct to the nearest degree. (5)
Mark scheme
Answer
Mark
Mark scheme
73
P1
for process to work with cosine eg \((DB^2 =)\ 12^2 + 9^2 - 2 \times 12 \times 9 \times \cos 60\ (= 117)\)
P1
for a correct order of operations to find DB, eg \(\sqrt{225 - 216 \times \cos 60}\) or \(\sqrt{\text{``}117\text{''}}\ (= 10.8\ldots)\)
P1
for working with sine rule and \(60^\circ\) eg \(\dfrac{\sin CBD}{9} = \dfrac{\sin 60}{[DB]}\) or \(\sin CBD = 0.720\ldots\) or \(CBD = 46.10\ldots\)
P1
for working with sine rule and \(125^\circ\) eg \(\dfrac{\sin ABD}{6} = \dfrac{\sin 125}{[DB]}\) or \(\sin ABD = 0.454\ldots\) or \(ABD = 27.02\ldots\)
A1
for answer within the range 72.5 to 73.5
Additional guidance
\(\sqrt{117} = 3\sqrt{13}\) Can be implied by correct use in subsequent calculations
3rd and 4th P1 can be awarded in any order [\(DB\)] must be clearly identified but may not be correct, but cannot be 6 or 9 or 12 \(CBD\) and \(ABD\) may be given as a variable, award marks unless contradicted
20 \(VABC\) is a solid pyramid. \(ABC\) is an equilateral triangle.
\(M\) is the midpoint of \(AB\). \(F\) is the point on \(MC\) such that \(MF : FC = 1 : 2\)
The vertex \(V\) is vertically above \(F\). \(VA = VB = VC\)
\(VF = 8\) cm Angle \(VCM = 52^\circ\)
Work out the side length of the equilateral triangle \(ABC\). Give your answer correct to 1 decimal place. (3)
Mark scheme
Answer
Mark
Mark scheme
10.8
P1
for process to find \(FC\), eg \(\tan 52 = \dfrac{8}{FC}\) (\(FC = 6.25(028..)\))
P1
for process that will lead to side length of \(ABC\), eg \(\sin 60 = \dfrac{\text{``}6.25\text{''} \times 1.5}{BC}\) or \(\cos 30 = \dfrac{\text{``}6.25\text{''} \times 1.5}{BC}\) or \((\text{``}6.25\text{''} \times 1.5)^2 + (0.5x)^2 = x^2\) oe
Work out the area of triangle \(ABC\). Give your answer correct to 3 significant figures. (4)
Mark scheme
Answer
Mark
Mark scheme
56.0
P1
for a start to the process by correctly substituting into the cosine rule to find an angle eg, \(18.2^2 = 14.6^2 + 7.9^2 - 2 \times 14.6 \times 7.9 \times \cos A\)
P1
for rearranging to find \(\cos A\), eg \(\cos A = \dfrac{14.6^2 + 7.9^2 - 18.2^2}{2 \times 14.6 \times 7.9}\ (= -0.2413\ldots)\) or \(A = 103.965\ldots\)
P1
for process to find the area of triangle \(ABC\) eg Area \(= \frac{1}{2} \times 7.9 \times 14.6 \times \sin(\text{``}103.965\ldots\text{''})\) or Area \(= \frac{1}{2} \times 7.9 \times 14.6 \times \sin[A]\) Area \(= \frac{1}{2} \times 14.6 \times 18.2 \times \sin(\text{``}24.912\ldots\text{''})\) or Area \(= \frac{1}{2} \times 14.6 \times 18.2 \times \sin[B]\) Area \(= \frac{1}{2} \times 7.9 \times 18.2 \times \sin(\text{``}51.1222\ldots\text{''})\) or Area \(= \frac{1}{2} \times 7.9 \times 18.2 \times \sin[C]\)
A1
for answer in the range 55.96 to 56.0
Additional guidance
\(\cos B = 0.9069\ldots\) \(B = 24.912\ldots\) \(\cos C = 0.6276\ldots\) \(C = 51.1222\ldots\)
[\(A\)], [\(B\)], [\(C\)] must be a numerical value and clearly identified by labelling or on the diagram with no contradiction
If an answer is given in the range in working and then rounded incorrectly award full marks
Calculate the area of triangle \(ACD\). Give your answer correct to 3 significant figures. (5)
Mark scheme
Answer
Mark
Mark scheme
52.5
P1
for start of process to find length of \(AC\), eg \(\dfrac{1}{2} \times 17 \times AC \times \sin 35 = 54\) or \((AC =)\ 54 \div \left(\dfrac{1}{2} \times 17 \times \sin 35\right)\ (= 11.076\ldots)\)
P1
for start of process to find \(AD\) or \(CD\), eg \(\dfrac{AD}{\sin 48} = \dfrac{\text{``}11.076\text{''}}{\sin 57}\) oe or \(\dfrac{AD}{\sin 48} = \dfrac{[AC]}{\sin 57}\) oe or \(\dfrac{CD}{\sin \text{``}75\text{''}} = \dfrac{\text{``}11.076\text{''}}{\sin 57}\) oe or \(\dfrac{CD}{\sin \text{``}75\text{''}} = \dfrac{[AC]}{\sin 57}\) oe
P1
for complete process to find \(AD\) or \(CD\) \((AD =)\ \dfrac{\text{``}11.076\text{''}}{\sin 57} \times \sin 48\ (= 9.81\ldots)\) or \((AD =)\ \dfrac{[AC]}{\sin 57} \times \sin 48\) or \((CD =)\ \dfrac{\text{``}11.076\text{''}}{\sin 57} \times \sin \text{``}75\text{''}\ (= 12.7\ldots)\) or \((CD =)\ \dfrac{[AC]}{\sin 57} \times \sin \text{``}75\text{''}\)
P1
for process to find area of triangle \(ACD\), eg \(\dfrac{1}{2} \times \text{``}11.076\text{''} \times \text{``}12.7\text{''} \times \sin 48\) or \(\dfrac{1}{2} \times [AC] \times [CD] \times \sin 48\) or \(\dfrac{1}{2} \times \text{``}11.076\text{''} \times \text{``}9.81\text{''} \times \sin \text{``}75\text{''}\) or \(\dfrac{1}{2} \times [AC] \times [AD] \times \sin \text{``}75\text{''}\) or \(\dfrac{1}{2} \times \text{``}9.81\text{''} \times \text{``}12.7\text{''} \times \sin 57\) or \(\dfrac{1}{2} \times [AD] \times [CD] \times \sin 57\)
A1
for answer in the range 52.4 to 52.52
Additional guidance
Check diagram for working throughout
Throughout \(\text{``}75\text{''} = 180 - 48 - 57\) \([AC]\) must be a numerical value and clearly identified by labelling or on the diagram with no contradiction.
\([AC]\), \([AD]\), \([CD]\) must be numerical values and clearly identified by labelling or on the diagram with no contradiction.
If an answer is shown in the range in working and then incorrectly rounded award full marks
\(P\) is the point on \(AB\) such that \(AP : PB = 5 : 2\)
Calculate the size of the angle between \(TP\) and the base \(ABCD\) of the prism. Give your answer correct to 1 decimal place. (4)
Mark scheme
Answer
Mark
Mark scheme
12.2
P1
for process to find \(AP\), eg \(\tfrac{5}{7} \times 14.7\ (= 10.5)\) or for showing the required angle on a diagram eg with an arc
P1
for process to find \(MP\), eg \(\sqrt{\text{``}10.5\text{''}^2 + 1.9^2}\) oe \((= 10.6\ldots)\) or for a process to find \(TP\), eg \(\sqrt{\text{``}10.5\text{''}^2 + 1.9^2 + 2.3^2}\) oe \((= 10.9\ldots)\)
P1
for using appropriate trigonometry ratio to find angle \(TPM\), eg \(\tan TPM = \dfrac{2.3}{\text{``}10.6\ldots\text{''}}\) or \(\sin TPM = \dfrac{2.3}{\text{``}10.9\ldots\text{''}}\) or \(\cos TPM = \dfrac{\text{``}10.6\ldots\text{''}}{\text{``}10.9\ldots\text{''}}\)
A1
for answer in the range 12.1 to 12.2
Additional guidance
Check diagram for working
If an answer is shown in the range in working and then incorrectly rounded award full marks.
Work out the size of the angle between \(FC\) and the plane \(ABCD\). (2)
Mark scheme
Answer
Mark
Mark scheme
30
P1
for a correct trig statement using angle \(ACF\), eg \(\sin x = \dfrac{6.8}{13.6}\) or \(x = \sin^{-1}\left(\dfrac{6.8}{13.6}\right)\) or \(\sin x = 0.5\) or \(\dfrac{6.8}{\sin x} = \dfrac{13.6}{\sin 90}\) oe
Work out the value of \(\sin ABC\) Give your answer in the form \(\dfrac{m}{n}\) where \(m\) and \(n\) are integers. (4)
Mark scheme
Answer
Mark
Mark scheme
\(\dfrac{65}{214}\)
B1
for \(\sin 30 = 0.5\)
P1
for use of the sine rule with values substituted, eg \(\dfrac{6.5}{\sin ABC} = \dfrac{10.7}{\sin 30}\) oe
P1
for \((\sin ABC =)\ \dfrac{6.5 \times \sin 30}{10.7}\) oe or for a complete process to find \(\sin ABC\), eg \((\sin ABC =)\ \dfrac{6.5 \times [0.5]}{10.7}\) oe
A1
for \(\dfrac{65}{214}\) oe eg \(\dfrac{325}{1070}\)
Additional guidance
Answer of \(\dfrac{3.25}{10.7}\) or \(\dfrac{6.5}{21.4}\) gets 3 marks
Where [0.5] is their value of \(\sin 30\)
Answer must be in the form \(\dfrac{m}{n}\) where \(m\) and \(n\) are integers
Calculate the length of \(QS\). Give your answer correct to 3 significant figures. You must show all your working. (4)
Mark scheme
Answer
Mark
Mark scheme
7.63
M1
for process to use the cosine rule to find \(QR\), eg \((QR^2 =)\ 11^2 + 9.4^2 - 2 \times 11 \times 9.4 \cos(27)\)
M1
for correct order of operations, eg \(QR = \sqrt{209.36 - 206.8 \times \cos 27}\ \ (= 5(.009\ldots))\) or \(QR = \sqrt{25(.09...)}\) or \(\sqrt{25.1}\)
M1
(dep on M1) for process to use the sine rule, eg \(\dfrac{QS}{\sin 88} = \dfrac{[QR]}{\sin 41}\) oe or \(QS = \dfrac{[QR] \times \sin 88}{\sin 41}\ (= 7.631...)\) oe
A1
for answer in range 7.61 to 7.632
Additional guidance
[\(QR\)] could be written as “5.009…” or could be a different figure, as long as this is clearly associated with the side \(QR\)
If an answer is given in the range in working and then rounded incorrectly award full marks. Award 0 marks for a correct answer with no (or incorrect) supportive working
\(D\) is the point on \(BC\) such that angle \(BAD\) = angle \(DAC = x^\circ\)
Prove that \(\dfrac{AB}{BD} = \dfrac{AC}{DC}\) (4)
Mark scheme
Answer
Mark
Mark scheme
Proof (supported)
M1
for using the sine rule on triangle \(ABD\) or on triangle \(ADC\), to involve sides \(AB\), \(BD\), \(AC\), or \(DC\) eg \(\dfrac{AB}{\sin ADB} = \dfrac{BD}{\sin x}\) oe or \(\dfrac{AC}{\sin ADC} = \dfrac{DC}{\sin x}\) oe OR for an expression for the area of triangle \(ABD\) or for the area of triangle \(ADC\) eg \(\dfrac{1}{2}AB\,AD \sin x\) or \(\dfrac{1}{2}AD\,AC \sin x\) or \(\dfrac{1}{2}h\,BD\) or \(\dfrac{1}{2}h\,DC\)
M1
for using the sine rule on both triangle \(ABD\) and on triangle \(ADC\), to involve sides \(AB\), \(BD\), \(AC\), or \(DC\) eg \(\dfrac{AB}{\sin ADB} = \dfrac{BD}{\sin x}\) oe and \(\dfrac{AC}{\sin ADC} = \dfrac{DC}{\sin x}\) oe OR for two expressions for the area of either triangle \(ABD\) or for triangle \(ADC\) eg \(\dfrac{1}{2}AB\,AD \sin x\) and \(\dfrac{1}{2}h\,BD\) or \(\dfrac{1}{2}AD\,AC \sin x\) and \(\dfrac{1}{2}h\,DC\)
M1
for stating or showing \(\sin ADB = \sin ADC\), eg \(\sin y = \sin(180 - y)\) OR for using two expressions to form an equation eg \(\dfrac{\frac{1}{2}AB\,AD \sin x}{\frac{1}{2}AD\,AC \sin x} = \dfrac{\frac{1}{2}h\,BD}{\frac{1}{2}h\,DC}\) oe
C1
for a full method to arrive at the given answer
Additional guidance
Accept extra letters eg \(y\) shown on diagram for any angle used
23 The diagram shows the positions of three towns, Acton (\(A\)), Barston (\(B\)) and Chorlton (\(C\)).
Barston is 8 km from Acton on a bearing of 037° Chorlton is 9 km from Barston on a bearing of 150°
Find the bearing of Chorlton from Acton. Give your answer correct to 1 decimal place. You must show all your working. (5)
Mark scheme
Answer
Mark
Mark scheme
098.6
P1
for using bearings to determine \(ABC\) as 67°
P1
for using the cosine rule to find \(AC\) eg (\(AC^2 =\)) \(9^2 + 8^2 - 2 \times 9 \times 8 \times \cos[67]\) oe or \(AC = 9.4199\ldots\)
P1
(dep P1) for using the sine rule to find angle \(BAC\) eg \(\dfrac{9}{\sin BAC} = \dfrac{\text{``}9.42\text{''}}{\sin[67]}\) oe OR for using the cosine rule to find angle \(BAC\) eg \(9^2 = \text{``}9.42^2\text{''} + 8^2 - 2 \times \text{``}9.42\text{''} \times 8 \times \cos BAC\) oe
P1
for rearranging eg \(\sin BAC = 9 \times \dfrac{\sin[67]}{\text{``}9.42\text{''}}\) oe OR eg \(\cos BAC = (\text{``}9.42^2\text{''} + 8^2 - 9^2) \div (2 \times \text{``}9.42\text{''} \times 8)\) oe OR for angle \(BAC = 61.57\ldots\)
A1
for angle in the range 98.5 to 98.6
Additional guidance
Accept 67 written on the diagram.
Accept correct substitution into RHS of equation Accept \(AC\) in the range 9.41 to 9.42
Accept any equivalent form with values substituted
If the correct answer is given without supportive evidence award 0 marks. Condone missing “0” at the front. If an answer within the range is seen in working and rounded incorrectly award full marks.
The base, \(ABCD\), of the prism is a square of side length 15 cm. Angle \(ABE\) and angle \(CBE\) are right angles. Angle \(EAB = 35^\circ\)
\(M\) is the point on \(DA\) such that
\(DM : MA = 2 : 3\)
Calculate the size of the angle between \(EM\) and the base of the prism. Give your answer correct to 1 decimal place. (4)
Mark scheme
Answer
Mark
Mark scheme
31.0
P1
for \(\tan 35 = BE \div 15\) or \(BE = 10.5(0\ldots)\) OR finding the length \(DM = \dfrac{2}{5} \times 15\ (= 6)\) or \(MA = \dfrac{3}{5} \times 15\ (= 9)\) or 6 : 9 OR showing the required angle on a diagram, eg with an arc
P1
for \(MB = \sqrt{15^2 + \text{``}9\text{''}^2}\) or \(\sqrt{306}\) or 17.4(9…) OR \(ME = \sqrt{\text{``}9\text{''}^2 + \text{``}18.3(1\ldots)\text{''}^2}\) or \(\sqrt{416.(3\ldots)}\) or 20.4(0…)
P1
for using an appropriate trigonometric ratio to set up an equation in angle \(EMB\) eg \(\tan\theta = \text{``}10.5(0\ldots)\text{''} \div \text{``}17.4(9\ldots)\text{''}\) or \(\cos\theta = \text{``}17.4(9\ldots)\text{''} \div \text{``}20.4(0\ldots)\text{''}\) or \(\sin\theta = \text{``}10.5(0\ldots)\text{''} \div \text{``}20.4(0\ldots)\text{''}\)
14 The diagram shows a rectangle, \(ABDE\), and two congruent triangles, \(AFE\) and \(BCD\).
area of rectangle \(ABDE\) = area of triangle \(AFE\) + area of triangle \(BCD\)
\(AB : AE = 1 : 3\)
Work out the length of \(AE\). (4)
Mark scheme
Answer
Mark
Mark scheme
36
P1
for process to find an expression for the area of triangle eg \(\tfrac{1}{2} \times 24 \times AE \times \sin 30\ (= 6AE)\)
P1
(dep P1) for process to link the area of rectangle with the area of the triangle eg \(2 \times \tfrac{1}{2} \times 24 \times AE \times \sin 30\ (= 12AE)\) or for \(AB = 12\)
P1
(indep) for use of given ratio eg \(AE = 3AB\) oe, eg area of rectangle = \(AE \times AB = 3x \times x\)
A1
cao
Additional guidance
Accept any correct expression, eg \(\tfrac{1}{2} \times 24 \times y \times \sin 30\)
May be shown on the diagram by labelling \(AE\) and \(AB\) with, for example, \(3x\), \(x\) or \(x\), \(\tfrac{1}{3}x\) or \(\tfrac{3}{4}x\), \(\tfrac{1}{4}x\) Do not accept 3, 1 or 1, \(\tfrac{1}{3}\) or \(\tfrac{3}{4}\), \(\tfrac{1}{4}\) for this mark.
Calculate the perimeter of the shape. Give your answer correct to 3 significant figures. (5)
Mark scheme
Answer
Mark
Mark scheme
39.9
P1
for finding the length of the minor or major arc eg \(\dfrac{220}{360} \times \pi \times 12\) (= 23(.03834..))
P1
for substituting into the sine or cosine rule to find \(OD\) eg \(14 \div \sin 140 = OD \div \sin 24\) or (\(OD^2 =\)) \(6^2 + 14^2 - 2 \times 6 \times 14 \times \cos 24\) (=78.5....)
P1
for a complete process to find the length \(OD\) eg \(14 \div \sin 140 \times \sin 24\) (=8.8(58778..))
P1
for a complete process to find the perimeter eg \(\text{``}23(.03834..)\text{''} + 14 + \text{``}8.8(58778..)\text{''} - 6\)
A1
for an answer in the range 39.8 to 40
Additional guidance
Allow appropriate rounding if calculation seen in parts
Must involve \(OD\) in the relationship but may be implied
May be seen in multiple calculations
If an answer in the range is seen in working and then incorrectly rounded award full marks.
13 The diagram shows a circle and an equilateral triangle.
One side of the equilateral triangle is a diameter of the circle. The circle has a circumference of 44 cm.
Work out the area of the triangle. Give your answer correct to 3 significant figures. (3)
Mark scheme
Answer
Mark
Mark scheme
84.9
P1
shows a process to find the radius or diameter eg \(44 = 2 \times \pi \times r\) or \(r = \dfrac{22}{\pi}\) or \(d = \dfrac{44}{\pi}\) or \(r = 7.0028\) or \(d = 14.0056..\)
P1
(dep on P1) complete method to find the area eg \(\frac{1}{2} \times \text{``}d\text{''}^2 \times \sin 60\) oe, \(\frac{1}{2} \times 14 \times \tan 60\) oe, \(\frac{1}{2} \times 14 \times \sqrt{14^2 - 7^2}\) oe
A1
for answer in the range 84.8 to 85
Additional guidance
Allow \(r\) in the range 7 to 7.1 and \(d\) in the range 14 to 14.1 Could be shown on the diagram.
If the correct answer in the range is given in working and then rounded incorrectly award full marks.
\(AB = 5\) m The vertex \(T\) is 12 m vertically above the midpoint of \(AC\).
Calculate the size of angle \(TAC\). (4)
Mark scheme
Answer
Mark
Mark scheme
73.6
P1
for correct initial use of Pythagoras eg \(5^2 + 5^2\) (=50) or a trigonometric ratio in the form \(\dfrac{5 \div 2}{0.5AC} = \sin 45\) oe
P1
for finding the length of half of the diagonal eg \(\sqrt{\text{``}50\text{''}} \div 2\) ( = 3.5...) or \(0.5AC = \dfrac{5 \div 2}{\sin 45}\) (=3.5...) oe
P1
for process to use tan eg \(\tan TAC = (12 \div \text{``}3.5..\text{''})\) (=3.3..) or complete alternative method arriving at an equation with the subject as \(\sin TAC\) or \(\cos TAC\)
A1
for an answer in the range 73.58 to 74.1
Additional guidance
P1 (half of the diagonal): do not accept \(\sqrt{20} \div 2\)
\(AB = 7.3\) 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)
Mark scheme
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
Additional guidance
P1 (start): Must include correct values
If an answer is given in the range but then incorrectly rounded award full marks.
Work out the length of \(AD\). Give your answer correct to 3 significant figures. (5)
Mark scheme
Answer
Mark
Mark scheme
13.1
P1
for start of process to find the length of \(BD\), eg \(\dfrac{BD}{\sin 34^\circ} = \dfrac{12.5}{\sin 109^\circ}\)
P1
for complete process to find the length of \(BD\), eg \(BD = \dfrac{12.5}{\sin 109^\circ} \times \sin 34^\circ\) (= 7.39...)
P1
for process to find the length of \(AD\), eg \(AD^2 = 11.4^2 + \text{``}7.39^2\text{''} - 2 \times 11.4 \times \text{``}7.39\text{''} \times \cos 86^\circ\)
P1
for process to use correct order of operations, eg \(129.96 + 54.6(5...) - 11.7(5...)\) (= 172.85...)
A1
for answer in the range 13.1 to 13.2
Additional guidance
Accept 7.4 for the award of the first two P marks
If an answer is given within the range and then incorrectly rounded to 3 sig figs award full marks.
\(ABEF\) and \(CBED\) are congruent parallelograms where \(AB = BC = x\) cm. \(P\) is the point on \(AF\) and \(Q\) is the point on \(CD\) such that \(BP = BQ = 10\) cm.
\(ONQ\) is a sector of a circle with centre \(O\) and radius 11 cm.
\(A\) is the point on \(ON\) and \(B\) is the point on \(OQ\) such that \(AOB\) is an equilateral triangle of side 7 cm.
Calculate the area of the shaded region as a percentage of the area of the sector \(ONQ\). Give your answer correct to 1 decimal place. (5)
Mark scheme
Answer
Mark
Notes
66.5
B1
for recognising an angle of 60 at \(AOB\)
P1
for a process to find the area of the sector, e.g. \(\dfrac{\text{``}60\text{''}}{360} \times \pi \times 11^2\) \(\left(= 63.3\ldots \text{ or } \dfrac{121\pi}{6}\right)\)
P1
for a process to find the area of the triangle, e.g. \(\dfrac{1}{2} \times 7^2 \times \sin \text{``}60\text{''}\) \(\left(= 21.2\ldots \text{ or } \dfrac{49\sqrt{3}}{4}\right)\)
P1
for a process to find the required percentage, eg. \(\dfrac{\text{``}63.3\ldots\text{''} - \text{``}21.2\ldots\text{''}}{\text{``}63.3\ldots\text{''}} \times 100\)
Work out the size of angle \(ACB\). Give your answer correct to 3 significant figures. (6)
Mark scheme
Answer
Mark
Notes
41.6
6
M1 for \(0.5 \times 5.4 \times 7.3 \times \sin B = 19\) M1 for \((B =) \sin^{-1}\left(\dfrac{19}{0.5 \times 5.4 \times 7.3}\right)\) oe (= 74.6) M1 for \((AC^2 =) 5.4^2 + 7.3^2 - 2 \times 5.4 \times 7.3 \times \cos \text{``}74.6\text{''}\) M1 (dep) for correct order of evaluation or 61.(479…) M1 for \(\dfrac{\sin C}{5.4} = \dfrac{\sin \text{``}74.6\text{''}}{\text{``}7.84\text{''}}\) oe or \(0.5 \times 7.3 \times \text{``}7.84\text{''} \times \sin C = 19\) A1 for answer in the range 41.55 to 41.65 from correct working OR (with perpendicular from \(C\) meeting \(AB\) at a point \(X\)) M1 for \(0.5 \times 5.4 \times CX = 19\) M1 for \((CX =) \dfrac{19}{0.5 \times 5.4}\) (= 7.037...) M1 for \(BCX = \cos^{-1} \dfrac{\text{``}7.04\text{''}}{7.3}\) (= 15.425…) M1 for \((BX =) \sqrt{7.3^2 - \text{``}7.04\text{''}^2}\) (= 1.94…) M1 for \(ACX = \tan^{-1} \dfrac{5.4 - \text{``}1.94\text{''}}{\text{``}7.04\text{''}}\) (= 26.17…) A1 for answer in the range 41.55 to 41.65 from correct working
OR
(with perpendicular from \(A\) meeting \(BC\) at a point \(Y\)) M1 for \(AY = \dfrac{19}{7.3 \div 2} = 5.20547..\) M1 for \(BY = \sqrt{5.4^2 - \text{``}5.20547\text{''}^2}\) (=1.43630..) M1 for \(CY = 7.3 - \text{``}1.43630..\text{''}\) (= 5.86369 ...) M1 for \(\tan C = \dfrac{\text{``}5.20547..\text{''}}{\text{``}5.86369..\text{''}}\) (= 0.8877..) M1 for \(C = \tan^{-1} \text{``}0.8877..\text{''}\) A1 for answer in the range 41.55 to 41.65 from correct working
Work out the size of angle \(BAC\). Give your answer correct to 3 significant figures. You must show all your working. (6)
Mark scheme
Answer
Mark
Notes
80.4
6
M1 for \(0.5 \times 7 \times 8 \times \sin x = 18\) M1 (dep) for \((x =)\ \sin^{-1}\left(\dfrac{18}{0.5 \times 7 \times 8}\right)\) oe (= 40) M1 (dep on at least M1) for \((AC^2 =)\ 7^2 + 8^2 - 2 \times 7 \times 8 \times \cos \text{``}40\text{''}\) M1 (dep on previous M1) for correct order of evaluation or 27.2(03…) or 5.2(15…) M1 (dep) for \(\sin A = \dfrac{8 \times \sin \text{``}40\text{''}}{\text{``}5.2(15..)\text{''}}\) or \(\sin A = \dfrac{18}{0.5 \times 7 \times \text{``}5.2(15..)\text{''}}\) or \(\cos A = \dfrac{\text{``}5.2(15..)\text{''}^2 + 7^2 - 8^2}{2 \times \text{``}5.2(15..)\text{''} \times 7}\) A1 for answer in the range 80.3 to 80.4 from correct working
OR (with perpendicular from \(A\) meeting \(BC\) at a point \(X\))
M1 for \(0.5 \times 8 \times h = 18\) M1 (dep) for \((h =)\ \dfrac{18}{0.5 \times 8}\) (= 4.5) M1 (dep on at least M1) for \(BAX = \cos^{-1}\dfrac{\text{``}4.5\text{''}}{7}\) (= 49.99 …) M1 (dep) for \((BX =)\ \sqrt{7^2 - \text{``}4.5\text{''}^2}\) (= 5.3619…) M1 (dep) for \(CAX = \tan^{-1}\dfrac{\text{``}4.5\text{''}}{8 - \text{``}5.3619\text{''}}\) (= 30.38 …) A1 for answer in the range 80.3 to 80.4 from correct working
NB Similar method applies for use of perpendicular from \(C\) to \(AB\)
\(DC = 7\) cm \(CB = 5\) cm Angle \(ABC\) is \(40^\circ\)
Work out the area of the parallelogram. Give your answer correct to 1 decimal place. (3)
Mark scheme
Answer
Mark
Notes
22.5
3
M1 for \(\tfrac{1}{2} \times 7 \times 5 \times \sin 40\) or \(\tfrac{1}{2} \times 7 \times 5 \times \sin(180 - 40)\) M1 (dep M1) for doubling the area of the triangle A1 for 22.4 – 22.5
OR M1 for complete method to find height of parallelogram, eg \(5 \sin 40^\circ\) M1 (dep M1) for complete method to find the area of the parallelogram, eg \(7 \times 5 \sin 40^\circ\) A1 for 22.4 – 22.5
(a) Work out the area of triangle \(ABC\). Give your answer correct to 3 significant figures. (2)
(b) Work out the length of the side \(AB\). Give your answer correct to 3 significant figures. (3)
Mark scheme (a)
Answer
Mark
Notes
18.2
2
M1 for \(\frac{1}{2} \times 6 \times 7 \times \sin 60\) A1 for answer in range 18.1 to 18.2
Mark scheme (b)
Answer
Mark
Notes
6.56
3
M1 for \(6^2 + 7^2 - 2 \times 6 \times 7 \times \cos 60\) M1 for correct order of operation eg \(36 + 49 - 42\) (=43) A1 for answer in range 6.55 to 6.56
Calculate the length of \(PR\). Give your answer correct to 3 significant figures. (3)
Mark scheme
Answer
Mark
Notes
15.0
3
M1 for \(8^2 + 8^2 - 2 \times 8 \times 8 \times \cos 140\) M1 (dep) for correct order of evaluation or 226.(05...) A1 for answer in range 15.0 − 15.04 OR M1 for \(\dfrac{PR}{\sin 140} = \dfrac{8}{\sin\left(\left(\frac{180 - 140}{2}\right)\right)}\) M1 for \(PR = \dfrac{8}{\sin\left(\left(\frac{180 - 140}{2}\right)\right)} \times \sin 140\) A1 for answer in range 15.0 − 15.04 OR M1 for \(8 \times \sin 70\) or \(8 \times \cos 20\) M1 for \(2 \times 8 \times \sin 70\) or \(2 \times 8 \times \cos 20\) A1 for answer in range 15.0 − 15.04
M1 (\(A =\)) \(\frac{1}{2} \times x \times 2x \times \sin 30^\circ\) A1 \(A = x^2 \times 0.5\) or \(A = \dfrac{x^2}{2}\) C1 for completion with all steps shown OR M1 height \(= 2x\sin 30\) (\(= x\)) A1 \(A = x^2 \times 0.5\) or \(A = \dfrac{x^2}{2}\) C1 for completion with all steps shown OR M1 for height \(= x\sin 30\) (\(= \frac{x}{2}\)) A1 \(A = x^2 \times 0.5\) or \(A = \dfrac{x^2}{2}\) C1 for completion with all steps shown
\(0.5 \times 6 \times x \times \sin 127 = 26.355\) or \(26.355 \div 6 \div \sin 127 \times 2\) or \(26.355 \div 6 \div [0.7986, 0.8] \times 2\) or [10.98, 11.001]
M1
length BC oe any letter
their \([10.98, 11.001] \div 4.4\) or [2.495, 2.50023] or \(4.4 \div\) their [10.98, 11.001] or [0.39996, 0.40073]
M1dep
length scale factor oe implied by [2.3997, 2.405] for length of DF
\(6 \div\) their [2.495, 2.50023] or \(6 \times\) their [0.39996, 0.40073] or [2.3997, 2.405] and \(0.5 \times 4.4 \times\) their [2.3997, 2.405] \(\times \sin 127\) or \(0.5 \times 4.4 \times\) their [2.3997, 2.405] \(\times\) [0.7986, 0.8] or \(26.355 \div (\text{their } [2.495, 2.50023])^2\) or \(26.355 \times (\text{their } [0.39996, 0.40073])^2\)
M1dep
length DF and area oe
area ratio and area \(26.355 \div \dfrac{[10.98, 11.001]^2}{4.4^2}\) or \(26.355 \times \dfrac{4.4^2}{[10.98, 11.001]^2}\) scores M3
Use the sine rule to work out the size of angle \(x\). [3 marks]
Mark scheme
Answer
Mark
Comments
\(\dfrac{\sin x}{17} = \dfrac{\sin 64}{23}\) or \(\sin x = \dfrac{17 \sin 64}{23}\) or \(\sin x = \dfrac{15.279\ldots}{23}\) or \(\dfrac{\sin x}{17} = 0.039\ldots\) or \(\sin x = 0.66(4\ldots)\)
27 The vertices of a regular hexagon lie on a circle with centre \(O\) and radius 5 cm
Not drawn accurately
Work out the shaded area.
Give your answer in the form \(\quad \dfrac{a\pi - b\sqrt{c}}{12} \quad\) where \(a\), \(b\) and \(c\) are integers. [4 marks]
Mark scheme
Answer
Mark
Comments
\(5^2 \times \pi\) (\(\div 6\)) or \(25\pi\) (\(\div 6\))
M1
oe allow 3.14 or better for \(\pi\) throughout
\(\dfrac{1}{2} \times 5 \times 5 \times \sin 60\) or \(\dfrac{1}{2} \times 5 \times 2.5 \tan 60\) or \(\dfrac{25}{2} \times \dfrac{\sqrt{3}}{2}\)
M1
oe correct method to work out the area of the triangle or the area of the hexagon implied by \(75 \sin 60\) or \(37.5 \tan 60\) or \(\dfrac{75\sqrt{3}}{2}\) oe
\(\dfrac{25\pi}{6} - \dfrac{25\sqrt{3}}{4}\)
A1
oe eg \(\dfrac{1}{6}\left(25\pi - \dfrac{75\sqrt{3}}{2}\right)\) implied by correct answer
\(\dfrac{50\pi - 75\sqrt{3}}{12}\)
A1
oe in correct form eg \(\dfrac{50\pi - 15\sqrt{75}}{12}\)
Additional guidance
Using Pythagoras to work out the perpendicular height of the triangle may lead to an area of \(\dfrac{5\sqrt{18.75}}{2}\) for the triangle or \(15\sqrt{18.75}\) for the area of the hexagon
Give your answer to 3 significant figures. [4 marks]
Mark scheme
Answer
Mark
Comments
\(\dfrac{1}{2} \times 14 \times AC = 80.5\)
M1
oe eg \(7AC = 80.5\) any letter for \(AC\)
\(\dfrac{80.5 \times 2}{14}\) or \(\dfrac{161}{14}\) or 11.5
M1dep
oe eg \(\dfrac{80.5}{7}\) implies M2 may be seen on diagram
\(\dfrac{1}{2} \times 19 \times\) their \(11.5 \times \sin 36\) or 64.21... or 64.22 or 64
M1
oe 64.21... or 64.22 or 64 scores M3 if no incorrect formula used
64.2 with no incorrect formula used
A1
Additional guidance
Answer 64.2 with no incorrect working
M3A1
11.5 scores M2 even if not subsequently used
Answer 64.2 from using ‘\(bh\)’ and ‘\(ab\sin C\)’ (unless clear explanation that \(\dfrac{1}{2}\) has been cancelled in both area formulae) \(14 \times AC = 80.5\) \(\dfrac{80.5}{14} = 5.75\) \(19 \times 5.75 \times \sin 36\) 64.2
M0 M0 M0 A0
3rd M1 can be scored if they have a value for \(AC\) eg \(AC = 6\) (may be seen on diagram) \(\dfrac{1}{2} \times 19 \times 6 \times \sin 36 = 33.5\)
M0M0 M1A0
3rd M1 may be seen in stages eg1 \(11.5 \times \sin 36\) or [6.7, 6.8] \(\dfrac{1}{2} \times 19 \times\) [6.7, 6.8] eg2 19 sin 36 or [11.1, 11.2] \(\dfrac{11.5 \times [11.1, 11.2]}{2}\)
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
Correct use of Pythagoras’ theorem using a square root
Alternative method 2
\(\sin E = \dfrac{8}{17}\;\) or \(\;\cos A = \dfrac{8}{17}\) or \(E = 28.(\ldots)\) or \(A = 61.9(\ldots)\) or 62 and \(\cos 28.(\ldots) = \dfrac{EM}{17}\) or \(\tan 28.(\ldots) = \dfrac{8}{EM}\) or \(\sin 61.9(\ldots) = \dfrac{EM}{17}\) or \(\tan 61.9(\ldots) = \dfrac{EM}{8}\)
M1
\(17 \cos 28.(\ldots)\) or \(8 \div \tan 28.(\ldots)\) or \(17 \sin 61.9(\ldots)\) or \(8 \tan 61.9(\ldots)\)
A1
Additional guidance
8, 15, 17 on their own
M0A0
\(EM^2 = 289 - 64 = 225,\ EM = 15\)
M1A0
Mark scheme (b)
Answer
Mark
Comments
Alternative method 1
\(30^2 + (16 \div 2)^2\) or \(\;30^2 + 8^2\) or 964
M1
oe
\(\sqrt{\text{their } 964}\) or \(2\sqrt{241}\) or [31, 31.1]
M1dep
oe CM
\(\tan x = \dfrac{15}{\text{their } [31,\ 31.1]}\)
M1dep
oe \(\;\) eg \(\;90 - \tan^{-1} \dfrac{\text{their } [31,\ 31.1]}{15}\) dep on M1 M1
[25.7, 26]
A1
Alternative method 2
\(30^2 + 17^2\) \(\;\) or \(\;\) 1189
M1
oe
\(\sqrt{\text{their } 1189}\) or [34.4, 34.5]
M1dep
oe CE
\(\sin x = \dfrac{15}{\text{their } [34.4,\ 34.5]}\)
M1dep
oe \(\;\) eg \(\;90 - \cos^{-1} \dfrac{15}{\text{their } [34.4,\ 34.5]}\) or \(\dfrac{\sin x}{15} = \dfrac{\sin 90}{\text{their } [34.4,\ 34.5]}\) dep on M1 M1
[25.7, 26]
A1
Alternative method 3
\(30^2 + (16 \div 2)^2\) or 964 or \(\;30^2 + 17^2\) or 1189
M1
oe
\(\sqrt{\text{their } 964}\) or \(2\sqrt{241}\) or [31, 31.1] or \(\sqrt{\text{their } 1189}\) or [34.4, 34.5]
M1dep
oe CM CE
\(\cos x = \dfrac{\text{their } [31,\ 31.1]}{\text{their } [34.4,\ 34.5]}\)
M1dep
oe \(\;\) eg \(\;90 - \sin^{-1} \dfrac{\text{their } [31,\ 31.1]}{\text{their } [34.4,\ 34.5]}\) dep on M1 M1
[25.7, 26]
A1
Alternative method 4
\(17^2 - (16 \div 2)^2\) or 225 or \(30^2 + (16 \div 2)^2\) or 964 or \(30^2 + 17^2\) or 1189
If (sector) 270 and (2 triangles) 240 followed by \(270 + 240 = 510\)
M4A1
Working back from 510. Apply scheme but maximum mark is M4A0
Assuming angle AEB = 72 and then using sine rule to work out BE does lead to area = 510 to 2sf but can score a maximum of M0M1M0M1depA0 \(BE = \dfrac{26}{\sin 72} \times \sin 38 = 16.8\) (or 17) \(\dfrac{108}{360} \times \pi \times 16.8^2 = 266 \qquad 2 \times \dfrac{1}{2} \times 15 \times 26 \times \sin 38 = 240.2\) \(506.2 \to 510\)
M0 M1M0depM1 A0
BE = [16.9, 17] seen with no working scores first M1 (and possibly all other marks)
\(BE = 35 \div 2 = 17.5 \to 17\) does not score first M1
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]
Mark scheme
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…
Appendix: Question 16
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
16 Work out the exact value of \(x\) in this triangle. [4]
Not to scale
Mark scheme
Answer
Marks
Part marks and guidance
\(\sqrt{31}\) final answer
4
B3 for answer 31 nfww
OR
B1 for \(\cos 30 = \frac{\sqrt{3}}{2}\)
and
M2 for \(x^2 = (6\sqrt{3})^2 + 7^2 - 2 \times 6\sqrt{3} \times 7\cos 30\) or better or M1 for other correct arrangements of cosine rule where \(x\) or \(x^2\) are not the subject
Do not accept work in decimals rounded to answers of \(\sqrt{31}\) or 31
Award if seen as clear statement of fact or in use; once seen condone replacement by 0.866… or 0.87 in subsequent work Condone missing brackets Condone use of 10.39… or 10.4 and 0.866… or 0.87 in M2 and M1
14 The diagram shows a circle, centre O and radius 52 cm.
Not to scale
AB is a chord of the circle. The line through T is a tangent to the circle.
The chord is parallel to the tangent. The perpendicular distance between the chord and the tangent is 72 cm.
Calculate the area of the shaded segment. You must show your working. [6]
Mark scheme
Answer
Marks
Part marks and guidance
2219.9… or 2220 with correct working
6
“Correct working” requires evidence of at least M2 or M1 M1 Accept answers in the range 2217 to 2226. Accept any correct method.
M2 for \(\cos [\ldots] = \frac{72 - 52}{52}\) oe or M1 for 72 – 52 or 20
and
M2 implied by 67.38… or 134.76… rot to at least an integer e.g. 67, 67.3, 67.4, 67.38, 134, equivalent method includes \(\cos [\ldots] = \frac{52^2 + 52^2 - 96^2}{2 \times 52 \times 52}\) condone with 72 instead of 96
M2 for \(\frac{1}{2} \times 52 \times 52 \times \sin\) their 134.76… oe or M1 for \(\sqrt{52^2 - (\textit{their } 20)^2}\) oe or 48 and M1 for \(\frac{1}{2} \times\) their96 × their20 oe
and
M2 for finding the area of the triangle OAB implied by 960 their 134.76… must come from a correct attempt to find angle AOB and condone using AB = 72 for M2 and M1 including \(\sqrt{52^2 - 36^2}\) for M1
M1 for \(\frac{\textit{their angle AOB}}{360} \times \pi \times 52^2\) or B1 for \(\pi \times 52^2\)
M1 implied by 3176 to 3186 or B1 implied by 2704\(\pi\), 8490 to 8496
If 0, 1 or 2 scored award SC3 for the correct answer with insufficient working If 0 or 1 scored award SC2 for 3179.918… or 5314.718.. rot to at least an integer with insufficient working
alternative method using major sector: M2 for angle as before implied by 225, 225.2, 225.23 or 225.24 M2 for area of triangle as before or M1 for the third side e.g. 48 M1 for \(\left[\frac{\textit{their reflex angle AOB}}{360}\right] \times \pi \times 52^2\) or 5314.718..
19 The diagram shows the positions of three towns A, B and C.
Not to scale
The bearing of town B from town A is 053°. The bearing of town C from town B is 108°. AB = 35 km and BC = 61 km.
Calculate AC. You must show your working. [5]
Mark scheme
Answer
Marks
Part marks and guidance
85.99… to 86 with correct working
5
B4 for 7395[…] with correct working
“Correct working” requires at least M1 from cosine rule
OR M2 for correct method to find angle 125 e.g. 53 + 180 – 108 or 360 – 108 – (180 – 53) or M1 for correct method to find angle 72 or 127 e.g. 180 – 108 or 180 – 53 and M2 for \(\sqrt{35^2 + 61^2 - 2 \times 35 \times 61 \times \cos(\textit{their } 125)}\) or M1 for [AC2 =] 352 + 612 – 2×35×61×cos(their 125)
M2 implied by [angle ABC =] 125
If 0, 1 or 2 marks scored, instead award SC3 for answer 85.99… to 86 with no or insufficient working If 0 or 1 mark scored, instead award SC2 for answer 7395[…] with no or insufficient working
18 A, B and C are points on the circumference of a circle.
Not to scale
Points C and D lie on a tangent to the circle at C. Angle BCD = 45°. Angle ACB = 58°. AB = 8 cm.
Find the length of BC. [4]
Mark scheme
Answer
Marks
Part marks and guidance
6.67 or 6.670 to 6.671
4
B1 for [BAC =] 45 soi
Accept 6.7 with working
AND
M2 for [BC=] \(\dfrac{8 \times \sin(45 \text{ or } \textit{their}\ \text{BAC})}{\sin 58}\) or M1 for \(\dfrac{BC}{\sin(45 \text{ or } \textit{their}\ \text{BAC})} = \dfrac{8}{\sin 58}\) oe
May be seen on diagram or within M2/M1 expressions
PQ = 9 cm, QR = 8 cm and RS = 13 cm. Angle QPS = 30° and angle PSQ = 15°.
Calculate angle QSR. Give your answer correct to 1 decimal place. You must show your working. [6]
Mark scheme
Answer
Marks
Part marks and guidance
Accept answers that round to 25.7 with correct working
6
M2 for \(\frac{9 \times \sin 30}{\sin 15}\) oe or M1 for \(\frac{9}{\sin 15} = \frac{[\ldots]}{\sin 30}\) oe
AND
M3 for correct cos rule with cos as the subject \(\frac{13^2 + (their\ 17.386\ldots)^2 - 8^2}{2 \times 13 \times (their\ 17.386\ldots)}\) oe or M2 for the above (M3) with one error or wrong angle selected or for \(8^2 = 13^2 + (their\ 17.386\ldots)^2 - 2 \times 13 \times (their\ 17.386\ldots) \times \cos[\ldots]\) or M1 for the above (M2) with one error or wrong angle selected
If 0 or 1 or 2 scored, instead award SC3 for 25.7 (or an answer rounding to 25.7) with no working or insufficient working
If 0 or 1 scored, instead award SC2 for 0.898 to 0.902 with no working or insufficient working
If 0 scored, instead award SC1 for [QS =] 17.3 to 17.4 with no working
“Correct working” requires evidence of at least M1 AND M2
(17.386…)
(0.900…)
e.g. for \(\frac{8^2 + (their\ 17.386\ldots)^2 - 13^2}{2 \times 8 \times (their\ 17.386\ldots)}\)
e.g. for \(13^2 = 8^2 + (their\ 17.386\ldots)^2 - 2 \times 8 \times (their\ 17.386\ldots) \times \cos[\ldots]\)
21 The diagram shows triangle ABC. X lies on BC such that angle AXC = 90°.
Not to scale
BC = 7.5 cm, angle ABC = 32° and angle ACB = 43°.
Work out length AX. You must show your working. [6]
Mark scheme
Answer
Marks
Part marks and guidance
2.8 to 2.81 with correct working
6
First M1 may be on diagram or within other M marks M1 for BAC = 180 − (32 + 43) or 58 + 47 or 105
All methods: “correct working” requires at least M1ANDM1ANDM1
First M1 may be on diagram or within other M marks M1 for BAC = 180 − (32 + 43) or 58 + 47 or 105
AND M2 for AC = \(\dfrac{7.5 \times \sin 32}{\sin(\textit{their } 105)}\) or M1 for \(\dfrac{\text{AC}}{\sin 32} = \dfrac{7.5}{\sin(\textit{their } 105)}\) oe
AND M2 for AB = \(\dfrac{7.5 \times \sin 43}{\sin(\textit{their } 105)}\) or M1 \(\dfrac{\text{AB}}{\sin 43} = \dfrac{7.5}{\sin(\textit{their } 105)}\) oe
AND A1 for AC = 4.11[4…] or 4.115 (M1A1 implies M2A1)
AND A1 for AB = 5.29[5…] or 5.3[0] (M1A1 implies M2A1)
AND M1 for \(\dfrac{\text{AX}}{\textit{their } 4.11} = \sin 43\) or their 4.11 × sin 43 oe
AND M1 for \(\dfrac{\text{AX}}{\textit{their } 5.3[0]} = \sin 32\) or their 5.3[0] × sin 32 oe
If 0, 1 or 2 scored, instead award SC3 for 2.8 to 2.81 with no or insufficient working If 0 or 1 scored, instead award SC2 for 4.11[4…], 4.115 with no or insufficient working
If 0, 1 or 2 scored, instead award SC3 for 2.8 to 2.81 with no or insufficient working If 0 or 1 scored, instead award SC2 for 5.29[5…] or 5.3[0] with no or insufficient working
Alternative Method splitting BC: First M1 may be on diagram or within other M marks. Accept other terms for \(d\). Do not accept numerical partition of BC.
M1 for BX = \(d\) and XC = \(7.5 - d\)
First M1 may be on diagram or within other M marks. Accept other terms for \(d\). Do not accept numerical partition of BC.
M1 BX = \(7.5 - d\) and XC = \(d\)
AND M2 for \(d = \dfrac{7.5\tan 43}{\tan 32 + \tan 43}\) oe or M1 for \(d\tan 32 = (7.5 - d)\tan 43\) oe
AND M2 for \(d = \dfrac{7.5\tan 32}{\tan 43 + \tan 32}\) oe or M1 for \(d\tan 43 = (7.5 - d)\tan 32\) oe
AND A1 for \(d\) = 4.49 to 4.5 (M1A1 implies M2A1)
AND A1 for \(d\) = 3 to 3.01 (M1A1 implies M2A1)
AND M1 for \(\dfrac{\text{AX}}{\textit{their } 4.49} = \tan 32\) or their 4.49 × tan 32 or for \(\dfrac{\text{AX}}{7.5 - \textit{their } 4.49} = \tan 43\) or (7.5 − their 4.49) × tan 43
AND M1 for \(\dfrac{\text{AX}}{\textit{their } 3.01} = \tan 43\) or their 3.01 × tan 43 or \(\dfrac{\text{AX}}{7.5 - \textit{their } 3.01} = \tan 32\) or (7.5 − their 3.01) × tan 32
If 0, 1 or 2 scored, instead award SC3 for 2.8 to 2.81 with no or insufficient working. If 0 or 1 scored, instead award SC2 for 4.49[…] or 4.5[0] with no or insufficient working.
If 0, 1 or 2 scored, instead award SC3 for 2.8 to 2.81 with no or insufficient working. If 0 or 1 scored, instead award SC2 for 4.49[…] or 4.5[0] with no or insufficient working.
Alternative Method using Areas The main scheme still applies. The first four marks will be identical and the area method eventually simplifies to the final M1 expression. This is just a much more complex method.
Calculate length AB. You must show your working. [6]
Mark scheme
Answer
Marks
Part marks and guidance
105 or 104.7 to 104.82
6
M2 for \(\dfrac{48 \times \sin 53}{85}\) oe implied by 26.8… or M1 for \(\dfrac{\sin 53}{85} = \dfrac{\sin[B]}{48}\) oe
M1 for [C=] 180 − 53 − their 26.807… or 100.19[…]
M2 for \(\dfrac{85 \times \sin \textit{their} C}{\sin 53}\) oe or M1 for \(\dfrac{[AB]}{\sin \textit{their} C} = \dfrac{85}{\sin 53}\) oe
“Correct working” requires evidence of at least M2 or M1M1
Alternative method cosine rule (AB = \(x\)) M3 for quadratic with coefficients evaluated or M2 for \(x^2 + (-2 \times 48 \times \cos 53)x + (48^2 - 85^2)\) [=0] oe or M1 for \(85^2 = x^2 + 48^2 - 2 \times x \times 48\cos 53\) AND M2 for correct use of quadratic formula or M1 for quadratic formula with at most one error
i.e. \(x^2 - 57.77\ldots x - 4921\) [= 0] oe
If 0 scored SC2 for 105 or 104.7 to 104.82 with no working or SC1 for 26.8… with no working
or M1 for \(\frac{1}{2} \times 80 \times 100 \times \sin 30\) oe
B1 for \(\sin 30 = \frac{1}{2}\) oe soi
Area of triangle = 2000 implies M1B1
Mark scheme (b)
Answer
Marks
Part marks and guidance
Conditions for growing may have been different in 2019 oe
1
e.g. extremes in weather oe disease in the carrots oe 2019 may not have been an “average” year oe 2019 may not have harvested the same number as other years Assumes the same amount will grow [in 2019]
18 The diagram shows a right-angled triangular prism ABCDEF.
Length AD = 11 cm, length CD = 10 cm and length CF = 6 cm.
(a) Calculate the volume of the prism. [2]
(b) Use trigonometry to show that angle FDC = 31°, correct to the nearest degree. [2]
(c) Calculate the exact length of AF. [4]
Mark scheme (a)
Answer
Marks
Part marks and guidance
330
2
M1 for \(\frac{1}{2} \times 10 \times 6 \times 11\)
Mark scheme (b)
Answer
Marks
Part marks and guidance
tan […] = \(\frac{6}{10}\) oe
1
accept any correct method, equivalents include tan-1(0.6)
30.9[…]
1
Mark scheme (c)
Answer
Marks
Part marks and guidance
\(\sqrt{257}\) cao
4
M3 for \(\sqrt{11^2 + 10^2 + 6^2}\) or better or M2 for 112 + 102 + 62 or M1 for 112 + 102 or 102 + 62 or [FD =] \(\frac{10}{\cos 31}\) or \(\frac{6}{\sin 31}\)
implied by 16.0[3…]
implied by 257
implied by [FD =] \(\sqrt{136}\) or 11.6 to 11.7 or [AC =] \(\sqrt{221}\) or 14.8 to 14.9
M2 for DF2 = 102 + 122 – 2 × 10 × 12 cos30 oe and allow one error or M1 for \(\cos 30 = \frac{10^2 + 12^2 - DF^2}{2 \times 10 \times 12}\), condone one error
nfww i.e. not 12 × sin 30 M2 implied by DF2 = 36.15 to 36.16 if they draw a perpendicular from F to P on DG award M1 for correct method to find both FP = 5 and PG= 8.6[6…] or 8.67 or 8.7 M1 for correct use of Pythagoras’ on triangle FPD
Mark scheme (b)
Answer
Marks
Part marks and guidance
57.3[…] or 57 nfww
4
M2 for [sin B = ] \(\frac{12.4 \times \sin 63}{12.8}\)
or M1 for \(\frac{\sin B}{12.4} = \frac{\sin 63}{12.8}\) oe and M1 for 180 – 63 – their 59.67
aircraft B is 4960 km from T on a bearing of 057°.
Not to scale
(a) Aircraft A flies directly towards radar tower T at a speed of 890 km/h.
At what time will the aircraft pass over radar tower T? Give your answer to the nearest minute. [4]
(b) Calculate the distance that was between aircraft A and aircraft B at 3pm. [4]
Mark scheme (a)
Answer
Marks
Part marks and guidance
6.39 [pm] or 1839
4
B3 for 39 or answer rounding to 39.1 or 3°39 to 3°39′6.07″ or 6°39 to 6°39′6.07″ or 219 or answer rounding to 219.1 OR M1 for [\(t =\)] \(\dfrac{3250}{890}\) oe soi by 3.65(…) and M1FT for 60 × (their time) soi or evidence from their answer by using calculator key
Alternative method (converting speed to km/min) M1 for 890 ÷ 60 soi by \(\dfrac{89}{6}\) or \(14\frac{5}{6}\) oe or 14.8[3…] and M1FT for [\(t =\)] 3250 ÷ their 14.8[3…]
Condone 1839pm for full marks
e.g. \(3\frac{58}{89}\)
their time could be fraction or decimal and could be just the non-integer part (check using calculator)
Mark scheme (b)
Answer
Marks
Part marks and guidance
3345 to 3350 nfww
4
B1 for 42 seen AND M2 for [\(x^2 =\)] \(3250^2 + 4960^2 - 2 \times 3250 \times 4960 \cos\theta\) oe soi by [\(x^2 =\)] 11205110 to 11205111 or M1 for correct cosine rule with \(x^2\) not as subject
Alternative method (using horizontal/vertical components and Pythagoras) M3 for \(\sqrt{(4960\sin 57 - 3250\sin 15)^2 + (3250\cos 15 - 4960\cos 57)^2}\) or M2 for 4960sin57 – 3250sin15 or 3250cos15 – 4960cos57 or M1 for two of 4960sin57, 3250sin15, 3250cos15 or 4960cos57
May be seen on sketch diagram
For M2 or M1, \(\theta\) is a number in the range \(15 \leqslant \theta \leqslant 57\)
20 The diagram shows a right-angled triangular prism ABCDEF.
Calculate angle AFB. [6]
Mark scheme
Answer
Marks
Part marks and guidance
20.9 or 20.89 or 20.886… or 21 with correct working
6
B1 for triangle AFB indicated e.g. drawn on diagram and M2 for [BF =] \(\sqrt{25^2 + 40^2}\) or 47.16[9…] or 47.17 or 47.2 or [AF =] \(\sqrt{25^2 + 40^2 + 18^2}\) or 50.48… or 50.5 or M1 for \(25^2 + 40^2\) [\(+\ 18^2\)] and M2 for e.g. \(\tan^{-1}\)(18 ÷ their 47.169…) or \(\sin^{-1}\)(18 ÷ their 50.48…) or M1 for [tan =] 18 ÷ their 47.169… or [sin =] 18 ÷ their 50.48…
their 47.169 should be an attempt at BF and their 50.48 should be an attempt at AF
M2 for \(x^2 - 10x + 19 = 0\) oe or M1 for \(9^2 = 10^2 + x^2 - 2 \times 10 \times x \times \cos 60\) AND M1FT for \(\dfrac{10 \pm \sqrt{10^2 - 4 \times 1 \times 19}}{2}\) A1 for \(x = 7.45\) or \(5 + \sqrt{6}\) AND M1 for \(\frac{1}{2} \times 10 \times \textit{their}\ 7.45 \times \sin 60\) oe
Alternative M1 for \(\dfrac{\sin 60}{9} = \dfrac{\sin B}{10}\) oe M1 for \(\sin B = \dfrac{10}{9}\sin 60\) or better A1 for B = 74.2(…) AND M1 for A = 180 – 60 – their 74.2 soi by 45.8 AND M1 for \(\frac{1}{2} \times 9 \times 10 \times \sin(\textit{their}\ 45.8)\)
Accept 32 after full correct method
Use of cosine rule
FT their quadratic = 0 Alternative: M1 for \((x - 5)^2 - 6 = 0\)
Ignore 2.55 or \(5 - \sqrt{6}\)
Their 7.45 should be from cosine rule followed by quadratic (not from measuring etc.)
Use of sine rule
Isolates sinB
Their 45.8 should be from sine rule followed by 180 – their sine rule answer (not from measuring etc.)
11.1 or 11.14 or 11.13[6…] or accept 11 with supporting working.
6
M3 for correct cos rule with cos as subject e.g. [cos = ] \(\dfrac{6.4^2 + 5.8^2 - 3.9^2}{2 \times 6.4 \times 5.8}\)
or M2 for the above (M3) formula with one error or for \(3.9^2 = 6.4^2 + 5.8^2 - 2 \times 6.4 \times 5.8 \times \cos[.]\) or M1 for this (M2) formula with one error AND M2 for \(\frac{1}{2} \times 6.4 \times 5.8 \times \sin(\textit{their}\,36.87\ldots)\) or M1 for the use of this formula with one error
accept any correct method and they can find any angle, see additional guidance for the other angles
this angle (opposite 3.9) is 36.87… which implies M3
M2 for \(\sqrt{11^2 + 6^2}\) soi by 12.52 to 12.53 or M1 for \([\ldots]^2 = 11^2 + 6^2\) and M2 for tan [… =] 8 ÷ their 12.52 or M1 for 8 ÷ their 12.52
accept any correct and full method note: HB = 14.866…
18 Alvin has a crate in the shape of a cuboid. The crate is open at the top. The internal dimensions of the crate are 46 cm long by 46 cm wide by 55 cm high.
Alvin has a stick of length 95 cm.
Alvin places the stick in the crate so that the shortest possible length extends out above the top of the crate.
(a) Calculate the length of the stick that extends out of the crate. [4]
(b) Calculate the angle the stick makes with the base of the crate. [3]
Mark scheme (a)
Answer
Marks
Part marks and guidance
9.8[1…] nfww
4
Accept answers rounding to 9.8 if correct working seen Condone for full marks minor inaccuracies from rounding, such as \(\sqrt{7256}\) seen
M3 for \(\sqrt{46^2 + 46^2 + 55^2}\) or 85.18 to 85.2 or \(\sqrt{7257}\)
OR
M2 for \(46^2 + 46^2 + 55^2\) or 7257 or \(\sqrt{46^2 + 46^2}\) or \(\sqrt{4232}\) or 65.05 to 65.1 or \(\sqrt{46^2 + 55^2}\) or \(\sqrt{5141}\) or 71.7[…]
OR
M1 for \(46^2 + 46^2\) or 4232 or \(46^2 + 55^2\) or 5141
May be done in steps
Mark scheme (b)
Answer
Marks
Part marks and guidance
40.2 nfww
3
M2 for sin […] = \(\dfrac{55}{\mathit{their}\,85.18 \text{ to } 85.2}\) or tan […] = \(\dfrac{55}{\mathit{their}\sqrt{46^2 + 46^2}}\) or cos […] = \(\dfrac{\mathit{their}\sqrt{46^2 + 46^2}}{\mathit{their}\,85.18 \text{ to } 85.2}\)
OR
M1 for indication of required angle
Accept 40° and answers rounding to 40.2 if correct working seen
0 for tan […] = \(\frac{55}{46}\) M2 for cosine rule with cos as subject
14 The diagram shows triangle ABC with D on AC and E on AB. DE is a straight line.
Not to scale
AD = 28 m, AE = 41 m, DE = 22 m and BC = 64 m.
Calculate the length CD. [6]
Mark scheme
Answer
Marks
Part marks and guidance
92 or 92.28 to 92.6
6
M3 for correct explicit cos rule to find angle A in ADE with cos as subject. [cos A =] \(\dfrac{28^2 + 41^2 - 22^2}{2 \times 28 \times 41}\) oe soi
accept any correct method
implied by [A=] 30.3 to 30.4
or
M2 for correct implicit form of the cos rule to find angle A \(22^2 = 28^2 + 41^2 - 2 \times 28 \times 41 \times \cos A\)
or
M1 for either of the above forms with only one error
AND
M2 for correct sine rule e.g. \(\dfrac{64 \times \sin 72}{\sin \mathit{their}A}\) oe soi or M1 for \(\dfrac{64}{\sin \mathit{their}A} = \dfrac{[\ldots]}{\sin 72}\) oe
if 0 scored SC1 for explicit form of cos rule to find angle D or E in ADE e.g. [cos D =] \(\dfrac{28^2 + 22^2 - 41^2}{2 \times 28 \times 22}\)
