Question Bank › IGCSE Shape & Space › Advanced Trigonometry
Advanced Trigonometry Topic Angles (Including Parallel Lines) (0) Angles in Polygons (2) Constructions & Bearings (3) Circle Theorems (3) Volume & Surface Area (4) Prisms (4) 2D Area & Perimeter (3) Circles & Sectors (3) Similar Shapes (4) Transformations of Shapes (2) Vectors (4) Basic Trigonometry (6) Advanced Trigonometry (7) Pythagoras (4) Plans, Elevations & Nets (0) Current PowerPoint version
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Higher June 2025 Paper 1R Q25
25
Diagram NOT accurately drawn
\(BAC\) is a sector of a circle, centre \(A\)
\(BCD\) is a sector of a circle, centre \(C\)
Angle \(BAC = 40^\circ\)2
Find the length of the arc \(BD\)
(6)
Mark scheme
Mark scheme Scheme Marks eg
\(\dfrac{40}{360}\pi r^2 - \dfrac{1}{2}r^2\sin 40 (= 28)\) oe or
\(\dfrac{40}{360}\pi r^2 = 28 + \dfrac{1}{2}r^2\sin 40\) oe
M1 (radius2 =) 992 – 1024 A1 eg \((BC^2 =)\;2 \times \text{``}{31.8}\text{''}^2 - 2 \times \text{``}{31.8}\text{''}^2\cos 40 (= 473.4\ldots)\)
or \(\dfrac{0.5BC}{\text{``}{31.8}\text{''}} = \sin 20\) or \(\dfrac{BC}{\sin 40} = \dfrac{\text{``}{31.8}\text{''}}{\sin(70)}\)
M1 eg \((BC =)\sqrt{2 \times \text{``}{31.8}\text{''}^2 - 2 \times \text{``}{31.8}\text{''}^2\cos 40}\;(= 21.7\ldots)\)
or \((BC =)\;2 \times \text{``}{31.8}\text{''}\sin 20 (= 21.7\ldots)\)
or \(BC = \dfrac{\text{``}{31.8}\text{''}\sin 40}{\sin(70)}(= 21.7\ldots)\)
M1 eg \(\dfrac{130}{360} \times 2 \times \pi \times \text{``}{21.7}\text{''}\) M1 Correct answer scores full marks (unless from obvious incorrect working) A1 (6) (6 marks)
Notes M1: for a correct expression for the area of the shaded region
Allow 3.14… or \(\dfrac{22}{7}\) for \(\pi\)
sin 40 = 0.64…
A1: Allow answers in the range 31.5 – 32.0
M1: for a correct first step to find \(BC\) using their clearly identified radius eg \(r\) = …. or seen on diagram
NB \(\dfrac{180 - 40}{2} = 70\)
sin 20 = 0.34…
sin 70 = 0.93… or 0.94
M1: dep on previous M1
M1: dep on previous M1
A1: accept 48.9 – 49.7
Higher June 2025 Paper 2 Q24
24 The diagram shows a square-based pyramid \(ABCDE\)
Diagram NOT accurately drawn
\(EA = EB = EC = ED\)
\(M\) is the centre of the horizontal square base \(ABCD\)
\(EA : AB = n : 1\)
Find the value of \(n\)
(4)
Mark scheme
Mark scheme Scheme Marks To find \(EQ\)
eg \(MQ = 0.5\) (= \(AQ\))
\(\dfrac{0.5}{\cos 80}\left(= 2.87(938..)\right)\)
eg \(MQ = x\) (= \(AQ\))
\(\dfrac{x}{\cos 80}(= 5.75..x)\)
NB cos80 = sin10
or To find \(EM\)
eg \(MQ = 0.5\) (= \(AQ\))
\(0.5\tan 80\left(= 2.83(564..)\right)\)
eg \(MQ = x\) (= \(AQ\))
\(x\tan 80(= 5.67..x)\)
NB \(\tan 80 = \dfrac{1}{\tan 10}\)
or To find \(EQ\)
eg \(EM = y\)
\(\dfrac{y}{\sin 80}(= 1.01..y)\)
To find \(MQ\) (= \(AQ\))
eg \(EQ = h\)
\(h\cos 80(= 0.173..h)\)
M1 To find \(EA^2\)
eg \(AQ = 0.5\) (= \(MQ\))
\(\text{``}{2.87..}\text{''}^2 + 0.5^2\)
\(\left(= 8.54(08..)\right)\)
eg \(AQ = x\) (= \(MQ\))
\(x^2 + \text{``}{\left(\dfrac{x}{\cos 80}\right)}\text{''}^2\)
or To find \(AM\)
eg \(MQ = 0.5\) (= \(AQ\))
\(\sqrt{0.5^2 + 0.5^2}\)
\(\left(= \dfrac{\sqrt{2}}{2} = 0.707(10..)\right)\)
eg \(MQ = x\) (= \(AQ\))
\(\sqrt{x^2 + x^2}\left(= \sqrt{2x^2} = x\sqrt{2}\right)\)
or To find \(MQ\) (= \(AQ\))
eg \(EM = y\)
\(\dfrac{y}{\tan 80}(= 0.176..y)\)
To find \(EA^2\)
eg \(EQ = h\)
\(h^2 + \text{``}{(h\cos 80)}\text{''}^2\)
M1 To find \(EA\)
eg \(AQ = 0.5\) (= \(MQ\))
\(\sqrt{\text{``}{2.87..}\text{''}^2 + 0.5^2}\)
\(\left(= \sqrt{8.54(08..)}\right)\) or
eg \(AQ = x\) (= \(MQ\))
\(\sqrt{x^2 + \text{``}{\left(\dfrac{x}{\cos 80}\right)}\text{''}^2}\;(= 5.84..x)\)
or To find \(EA\)
eg \(AQ = 0.5\) (= \(MQ\))
\(\sqrt{\text{``}{2.83..}\text{''}^2 + \text{``}{0.707..}\text{''}^2}\)
\(\left(= \sqrt{8.54(08..)}\right)\)
eg \(AQ = x\) (= \(MQ\))
\(\sqrt{\text{``}{(x\tan 80)}\text{''}^2 + \text{``}{(x\sqrt{2})}\text{''}^2}\)
\((= 5.84..x)\)
or To find \(EA\)
eg \(EM = y\)
\(\sqrt{\text{``}{\left(\dfrac{1}{\sin 80}\right)}\text{''}^2 + \text{``}{\left(\dfrac{1}{\tan 80}\right)}\text{''}^2}\)
\((= 1.03..y)\)
[= 1.03.. × \(AQ\) × tan80]
eg \(EQ = h\)
\(\sqrt{h^2 + \text{``}{(h\cos 80)}\text{''}^2}\;(= 1.01..h)\)
\(\left[= 1.01.. \times \dfrac{0.5}{\cos 80}\right]\)
M1 Working not required, so correct answer scores full marks (unless from obvious incorrect working) A1 (4) (4 marks)
Notes M1: Use of a value for the side \(AB\) eg 1 or \(2x\) or any value etc or Let \(EM = y\) or any value or \(EQ = h\)
A1: awrt 2.92
Higher June 2025 Paper 2R Q23
23 The diagram shows a triangular prism, \(ABCDEF\), with a horizontal rectangular base \(ABCD\)
Diagram NOT accurately drawn
\(M\) is the midpoint of the line \(AC\)
\(AC = 40\) cm angle \(CAE = 35^\circ\) angle \(ABF = 90^\circ\)
Work out the size of angle \(CME\)
(3)
Mark scheme
Mark scheme Scheme Marks eg
\((CE =)\;40 \times \tan 35\;(= 28(.0\ldots))\)
or \((CE =)\;\dfrac{40}{\tan(90 - 35)}\;(= 28(.0\ldots))\)
or \((CE =)\;\dfrac{40\sin 35}{\sin(90 - 35)}\;(= 28(.0\ldots))\)
M1 eg
(angle \(CME\) =) \(\tan^{-1}\left(\dfrac{\text{``}{28}\text{''}}{40 \div 2}\right)\)
or (angle \(CME\) =) \(\sin^{-1}\left(\dfrac{\text{``}{28}\text{''}}{\sqrt{(40 \div 2)^2 + \text{``}{28}\text{''}^2}}\right)\)
or (angle \(CME\) =) \(\cos^{-1}\left(\dfrac{40 \div 2}{\sqrt{(40 \div 2)^2 + \text{``}{28}\text{''}^2}}\right)\)
M1 Correct answer only scores full marks (unless from obviously incorrect working) A1 (3) (3 marks)
Notes M1: for a correct method to find \(CE\)
M1: for a complete method to find angle \(CME\)
A1: awrt 54.5
Higher June 2025 Paper 1 Q20
20
Diagram NOT accurately drawn
Work out the length of \(BC\)
(5)
Mark scheme
Mark scheme Scheme Marks \(\left(BD^2 =\right)9.4^2 + 12.8^2 - 2 \times 9.4 \times 12.8 \times \cos 72\) (= 177.8…)
or \(\left(BD^2 =\right)88.36 + 163.84 - 2 \times 9.4 \times 12.8 \times \cos 72\) (=177.8…) oe
or \(\left(BD =\right)\sqrt{9.4^2 + 12.8^2 - 2 \times 9.4 \times 12.8 \times \cos 72}\)
M1 (\(BD\) = )13.3 A1 eg \(\dfrac{BC}{\sin 39} = \dfrac{\text{``}{13.3\ldots}\text{''}}{\sin 54}\) or \(\dfrac{\sin 39}{BC} = \dfrac{\sin 54}{\text{``}{13.3\ldots}\text{''}}\) M1ft \(\left(BC =\right)\dfrac{\text{``}{13.3\ldots}\text{''}}{\sin 54} \times \sin 39\) M1ft Correct answer scores full marks (unless from obvious incorrect working) A1 (5) (5 marks)
Notes M1: for applying cosine rule
A1: allow 13.3 – 13.342 or \(\sqrt{177.8\ldots}\) or \(\sqrt{178}\)
M1ft: for applying the sine rule, allow use of their \(BD\)
M1ft: for method to find \(BC\) using the sine rule
A1: allow 10.3 – 10.4
Higher November 2024 Paper 2 Q24
24 \(A\), \(B\) and \(C\) are three points on horizontal ground.
\(AB = 8.4\) metres \(BC = 9.2\) metres
\(B\) is on a bearing of 067° from \(A\)
Calculate the bearing of \(A\) from \(C\)
(6)
Mark scheme
Mark scheme Scheme Marks 67 + 51 (= 118) or angle split into 67 and 51 M1 \(AC^2 = 8.4^2 + 9.2^2 - 2 \times 8.4 \times 9.2 \times \cos\text{``}{118}\text{''}\;(= 227.(7615))\) M1 \(AC = \sqrt{8.4^2 + 9.2^2 - 2 \times 8.4 \times 9.2 \times \cos\text{``}{118}\text{''}}\;(= 15.(09\ldots))\)
(15 – 15.1)
M1 \(\dfrac{\sin ACB}{8.4} = \dfrac{\sin\text{``}{118}\text{''}}{\text{``}{15.09\ldots}\text{''}}\) or
\(\dfrac{\sin BAC}{9.2} = \dfrac{\sin\text{``}{118}\text{''}}{\text{``}{15.09\ldots}\text{''}}\) or
\(\cos ACB = \dfrac{9.2^2 + \text{``}{15.09}\text{''}^2 - 8.4^2}{2 \times 9.2 \times \text{``}{15.09}\text{''}}\) or \(\cos BAC = \dfrac{8.4^2 + \text{``}{15.09}\text{''}^2 - 9.2^2}{2 \times 8.4 \times \text{``}{15.09}\text{''}}\)
M1 \((ACB =)\sin^{-1}\left(\dfrac{8.4\sin\text{``}{118}\text{''}}{\text{``}{15.09\ldots}\text{''}}\right)\) or \(\cos^{-1}\left(\dfrac{9.2^2 + \text{``}{15.09}\text{''}^2 - 8.4^2}{2 \times 9.2 \times \text{``}{15.09}\text{''}}\right)\;(= 29.(43\ldots))\)
(29 – 30)
\((BAC =)\sin^{-1}\left(\dfrac{9.2\sin\text{``}{118}\text{''}}{\text{``}{15.09\ldots}\text{''}}\right)\) or \(\cos^{-1}\left(\dfrac{8.4^2 + \text{``}{15.09}\text{''}^2 - 9.2^2}{2 \times 8.4 \times \text{``}{15.09}\text{''}}\right)\;(= 32.(56\ldots))\)
(32 – 33)
M1 Correct answer scores full marks (unless from obvious incorrect working) Look for angles and lengths on their diagram A1 (6) (6 marks)
Notes M1: A diagram showing 118 or use of 118 in further calculations
M1: a correct method to find length \(AC^2\)
M1: a correct method to find length \(AC\)
M1: dep on previous method marks
M1: a completely correct statement for angle \(ACB\) or angle \(BAC\)
A1: allow 279 – 280
Higher November 2024 Paper 1 Q16
16 \(OAPB\) is a sector of a circle, centre \(O\)
Diagram NOT accurately drawn
Angle \(AOB = 50^\circ\)
Area of triangle \(OAB = 120\) cm2
Work out the area of the sector \(OAPB\)
(4)
Mark scheme
Mark scheme Scheme Marks \(120 = \dfrac{1}{2} \times a \times b \times \sin 50\) oe or \(\dfrac{120 \times 2}{\sin 50}\) oe
or
\(120 = \dfrac{1}{2} \times 2x\sin 25 \times x\cos 25\) oe or \(\dfrac{120 \times 2}{2 \times \sin 25 \times \cos 25}\) oe
or
313(.2977494)
M1 \((\textit{radius} =)\sqrt{\dfrac{120 \times 2}{\sin 50}}\) (= 17.7(0021891)) oe or
\((\textit{radius} =)\sqrt{\dfrac{120 \times 2}{2 \times \sin 25 \times \cos 25}}\) (= 17.7(0021891)) oe or
\((\textit{radius} =)\sqrt{313(.2977494)}\) (= 17.7(0021891))
M1 (area of \(OAPB\) =) \(\pi \times \text{“}17.7\text{”}^2 \times \dfrac{50}{360}\) oe M1 Correct answer scores full marks (unless from obvious incorrect working) A1 (4) (4 marks)
Notes M1: for a correct equation for \(a \times b\) or \(r^2\) (allow any letters for \(a\) or \(b\))correct expression for \(a \times b\) or \(r^2\)
M1: for a correct rearrangement to find the radius
A1: awrt 137
Higher November 2024 Paper 1 Q12
12 \(ABC\) is a right-angled triangle.
Diagram NOT accurately drawn
\(AB = 4250\) m angle \(BAD = 47^\circ\) angle \(BCA = 24^\circ\)
Work out the length of \(DC\)
(4)
Mark scheme
Mark scheme Scheme Marks eg
\(\tan 47 = \dfrac{(BD)}{4250}\) or \(\tan 24 = \dfrac{4250}{(BC)}\) or \(\tan(47 + \text{“}19\text{”}) = \dfrac{(BC)}{4250}\) or \(\dfrac{(BD)}{\sin 47} = \dfrac{4250}{\sin 43}\)
or \((AD =)\;\dfrac{4250}{\cos 47}\;(= 6231.686\ldots)\) or \((AD =)\;\dfrac{4250}{\sin 43}\;(= 6231.686\ldots)\)
or \((AC =)\;\dfrac{4250}{\sin 24}\;(= 10\,449.021\ldots)\) or \((AC =)\;\dfrac{4250}{\cos 66}\;(= 10\,449.021\ldots)\)
M1 eg
\((BD =)\;4250\tan 47\;(= 4557.567\ldots)\) or \((BC =)\;\dfrac{4250}{\tan 24}\;(= 9545.656\ldots)\)
or \((BD =)\;\dfrac{4250}{\sin 43} \times \sin 47\;(= 4557.567\ldots)\) or \(\dfrac{(DC)}{\sin\text{“}19\text{”}} = \dfrac{\text{“}10449\ldots\ldots\text{”}}{\sin\text{“}137\text{”}}\)
or \(\dfrac{(DC)}{\sin\text{“}19\text{”}} = \dfrac{\text{“}6231.686\text{”}}{\sin 24}\) or \((BC =)\;4250 \times \tan(47 + \text{“}19\text{”})\;(= 9545.656\ldots)\)
or \((DC^2 =)\;\text{“}6231\text{”}^2 + \text{“}10449\text{”}^2 - 2 \times \text{“}6231\text{”} \times \text{“}10449\text{”} \times \cos 19\)
M1 eg
“9545.656” – “4557.567” (= 4988.089)
or \((DC =)\;\dfrac{\text{“}6231.686\text{”}}{\sin 24} \times \sin 19\) or \((DC =)\;\dfrac{\text{“}10449\ldots\ldots\text{”}}{\sin\text{“}137\text{”}} \times \sin\text{“}19\text{”}\)
or \((DC =)\sqrt{\text{“}6231\text{”}^2 + \text{“}10449\text{”}^2 - 2 \times \text{“}6231\text{”} \times \text{“}10449\text{”} \times \cos 19}\)
M1 Correct answer scores full marks (unless from obvious incorrect working) A1 (4) (4 marks)
Notes M1: for a complete method
A1: allow in the range 4932 – 4990
No questions match these filters.
