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Question
On a map three schools A, B and C are situated as shown in the diagram.
Schools A and B are 625 metres apart.
Angle ABC = 102° and BC = 986 metres.
Find the distance between A and C.[3]
Find the size of angle BAC.[3]
Answer/Explanation
Markscheme
Unit penalty (UP) is applicable in question part (a) only.
\({\text{AC}}^2 = 625^2 + 986^2 – 2 \times 625 \times 986 \times \cos102^\circ\) (M1)(A1)
( = 1619072.159)
AC = 1272.43
(UP) =1270 m (A1) (C3)[3 marks]
\(\frac{{986}}{{\operatorname{sinA} }} = \frac{{1270}}{{\sin 102^\circ }}\) (M1)(A1)(ft)
\({\text{A}} = 49.4^\circ\) (A1)(ft)
OR
\(\frac{{986}}{{\operatorname{sinA} }} = \frac{{1272.43}}{{\sin 102^\circ }}\) (M1)(A1)(ft)
\({\text{A}} = 49.3^\circ\) (A1)(ft)
OR
\(\cos {\text{A}} = \left( {\frac{{{{625}^2} + {{1270}^2} – {{986}^2}}}{{2 \times 625 \times 1270}}} \right)\) (M1)(A1)(ft)
\({\text{A}} = 49.5^\circ\) (A1)(ft) (C3)[3 marks]
Question
Triangle \({\text{ABC}}\) is such that \({\text{AC}}\) is \(7{\text{ cm}}\), angle \({\text{ABC}}\) is \({65^ \circ }\) and angle \({\text{ACB}}\) is \({30^ \circ }\).
Sketch the triangle writing in the side length and angles.[1]
Calculate the length of \({\text{AB}}\).[2]
Find the area of triangle \({\text{ABC}}\).[3]
Answer/Explanation
Markscheme
(A1) (C1)
Note: (A1) for fully labelled sketch.[1 mark]
Unit penalty (UP) may apply in this question.
\(\frac{{{\text{AB}}}}{{\sin 30}} = \frac{7}{{\sin 65}}\) (M1)
(UP) \({\text{AB}} = 3.86{\text{ cm}}\) (A1)(ft) (C2)
Note: (M1) for use of sine rule with correct values substituted.[2 marks]
Unit penalty (UP) may apply in this question.
\({\text{Angle BAC}} = {85^ \circ }\) (A1)
\({\text{Area}} = \frac{1}{2} \times 7 \times 3.86 \times \sin {85^ \circ }\) (M1)
(UP) \( = 13.5{\text{ }}{{\text{cm}}^2}\) (A1)(ft) (C3)[3 marks]
Question
The diagram shows triangle ABC in which angle BAC \( = 30^\circ \), BC \( = 6.7\) cm and AC \( = 13.4\) cm.
Calculate the size of angle ACB.[4]
Nadia makes an accurate drawing of triangle ABC. She measures angle BAC and finds it to be 29°.
Calculate the percentage error in Nadia’s measurement of angle BAC.[2]
Answer/Explanation
Markscheme
\(\frac{{\sin {\text{A}}{\operatorname{\hat B}}{\text{C}}}}{{13.4}} = \frac{{\sin 30^\circ }}{{6.7}}\) (M1)(A1)
Note: Award (M1) for correct substituted formula, (A1) for correct substitution.
\({\text{A}}{\operatorname{\hat B}}{\text{C}}\) = 90° (A1)
\({\text{A}}{\operatorname{\hat C}}{\text{B}}\) = 60° (A1)(ft) (C4)
Note: Radians give no solution, award maximum (M1)(A1)(A0).[4 marks]
\(\frac{{29 – 30}}{{30}} \times 100\) (M1)
Note: Award (M1) for correct substitution into correct formula.
% error = −33.3 % (A1) (C2)
Notes: Percentage symbol not required. Accept positive answer.[2 marks]
Question
The diagram shows a triangle ABC in which AC = 17 cm. M is the midpoint of AC.
Triangle ABM is equilateral.
Write down the size of angle MCB.[1]
Write down the length of BM in cm.[1]
Write down the size of angle BMC.[1]
Calculate the length of BC in cm.[3]
Answer/Explanation
Markscheme
30° (A1) (C3)[1 mark]
8.5 (cm) (A1)[1 mark]
120° (A1)[1 mark]
\(\frac{{{\text{BC}}}}{{\sin 120}} = \frac{{8.5}}{{\sin 30}}\) (M1)(A1)(ft)
Note: Award (M1) for correct substituted formula, (A1) for correct substitutions.
\({\text{BC}} = {\text{14}}{\text{.7}}\left( {\frac{{17\sqrt 3 }}{2}} \right)\) (A1)(ft)[3 marks]
Question
The base of a prism is a regular hexagon. The centre of the hexagon is O and the length of OA is 15 cm.
Write down the size of angle AOB.[1]
Find the area of the triangle AOB.[3]
The height of the prism is 20 cm.
Find the volume of the prism.[2]
Answer/Explanation
Markscheme
60° (A1) (C1)[1 mark]
\(\frac{{15 \times \sqrt {{{15}^2} – {{7.5}^2}} }}{2} = 97.4{\text{ c}}{{\text{m}}^2}\) (97.5 cm2) (A1)(M1)(A1)
Notes: Award (A1) for correct height, (M1) for substitution in the area formula, (A1) for correct answer.
Accept 97.5 cm2 from taking the height to be 13 cm.
OR
\(\frac{1}{2} \times {15^2} \times \sin 60^\circ = 97.4{\text{ c}}{{\text{m}}^2}\) (M1)(A1)(A1)(ft) (C3)
Notes: Award (M1) for substituted formula of the area of a triangle, (A1) for correct substitution, (A1)(ft) for answer.
Follow through from their answer to part (a).
If radians used award at most (M1)(A1)(A0).[3 marks]
97.4 × 120 = 11700 cm3 (M1)(A1)(ft) (C2)
Notes: Award (M1) for multiplying their part (b) by 120.[2 marks]
Question
In the diagram, \({\text{AD}} = 4{\text{ m}}\), \({\text{AB}} = 9{\text{ m}}\), \({\text{BC}} = 10{\text{ m}}\), \({\text{B}}\hat {\text{D}}{\text{A}} = {90^ \circ }\) and \({\text{D}}\hat {\text{B}}{\text{C}} = {100^ \circ }\) .
Calculate the size of \({\text{A}}\hat {\text{B}}{\text{C}}\).[3]
Calculate the length of AC.[3]
Answer/Explanation
Markscheme
\(\sin {\text{A}}\hat {\text{B}}{\text{D}} = \frac{4}{9}\) (M1)
\(100 + {\text{their }}({\text{A}}\hat {\text{B}}{\text{D}})\) (M1)
\(126\% \) (A1) (C3)
Notes: Accept an equivalent trigonometrical equation involving angle ABD for the first (M1).
Radians used gives \(100\% \) . Award at most (M1)(M1)(A0) if working shown.
\({\text{BD}} = 8{\text{ m}}\) leading to \(127\% \) . Award at most (M1)(M1)(A0) (premature rounding).[3 marks]
\({\text{A}}{{\text{C}}^2} = {10^2} + {9^2} – 2 \times 10 \times 9 \times \cos (126.38 \ldots )\) (M1)(A1)
Notes: Award (M1) for substituted cosine formula. Award (A1) for correct substitution using their answer to part (a).
\({\text{AC}} = 17.0{\text{ m}}\) (A1)(ft) (C3)
Notes: Accept \(16.9{\text{ m}}\) for using \(126\). Follow through from their answer to part (a). Radians used gives \(5.08\). Award at most (M1)(A1)(A0)(ft) if working shown.[3 marks]
Question
The diagram shows quadrilateral ABCD in which AB = 13 m , AD = 6 m and DC = 10 m. Angle ADC =120° and angle ABC = 40 °.
Calculate the length of AC.[3]
Calculate the size of angle ACB.[3]
Answer/Explanation
Markscheme
AC2 = 62 + 102 – 2×10×6×cos120° (M1)(A1)
Note: Award (M1) for substitution in cosine formula, (A1) for correct substitutions.
AC = 14 (m) (A1) (C3)[3 marks]
\(\frac{{14}}{{\sin 40}} = \frac{{13}}{{\operatorname{sinACB} }}\) (M1)(A1)(ft)
Note: Award (M1) for substitution in sine formula, (A1) for correct substitutions.
Angle ACB = 36.6° (36.6463…) (A1)(ft) (C3)
Note: Follow through from their (a).[3 marks]
Question
In triangle ABC, BC = 8 m, angle ACB = 110 °, angle CAB = 40°, and angle ABC = 30 °.
Find the length of AC.[3]
Find the area of triangle ABC.[3]
Answer/Explanation
Markscheme
\(\frac{{{\text{AC}}}}{{\sin 30^\circ }} = \frac{8}{{\sin 40^\circ }}\) (M1)(A1)
Note: Award (M1) for substitution in the sine rule formula, (A1) for correct substitutions.
AC = 6.22 (m) (6.22289…) (A1) (C3)[3 marks]
Area of triangle \({\text{ABC}} = \frac{1}{2} \times 8 \times 6.22289… \times \sin 110^\circ \) (M1)(A1)(ft)
Note: Award (M1) for substitution in the correct formula, (A1)(ft) for their correct substitutions. Follow through from their part (a).
Area triangle ABC = 23.4 m2 (23.3904…m2) (A1)(ft) (C3)
Note: Follow through from a positive answer to their part (a). The answer is 23.4 m2, units are required.[3 marks]
Question
In the diagram, triangle ABC is isosceles. AB = AC and angle ACB is 32°. The length of side AC is x cm.
Write down the size of angle CBA.[1]
Write down the size of angle CAB.[1]
The area of triangle ABC is 360 cm2. Calculate the length of side AC. Express your answer in millimetres.[4]
Answer/Explanation
Markscheme
32° (A1) (C1)[1 mark]
116° (A1) (C1)[1 mark]
\(360 = \frac{1}{2} \times {x^2} \times \sin 116^\circ \) (M1)(A1)(ft)
Notes: Award (M1) for substitution into correct formula with 360 seen, (A1)(ft) for correct substitution, follow through from their answer to part (b).
x = 28.3 (cm) (A1)(ft)
x = 283 (mm) (A1)(ft) (C4)
Notes: The final (A1)(ft) is for their cm answer converted to mm. If their incorrect cm answer is seen the final (A1)(ft) can be awarded for correct conversion to mm.[4 marks]
Question
The quadrilateral ABCD has AB = 10 cm, AD = 12 cm and CD = 7 cm.
The size of angle ABC is 100° and the size of angle ACB is 50°.
Find the length of AC in centimetres.[3]
Find the size of angle ADC.[3]
Answer/Explanation
Markscheme
\(\frac{{{\text{AC}}}}{{\sin 100^\circ }} = \frac{{10}}{{\sin 50^\circ }}\) (M1)(A1)
Note: Award (M1) for substitution in the sine rule formula, (A1) for correct substitutions.
\(=12.9 (12.8557…)\) (A1) (C3)
Note: Radian answer is 19.3, award (M1)(A1)(A0).
\(\frac{{{{12}^2} + {7^2} – {{12.8557…}^2}}}{{2 \times 12 \times 7}}\) (M1)(A1)(ft)
Note: Award (M1) for substitution in the cosine rule formula, (A1)(ft) for correct substitutions.
= 80.5° (80.4994…°) (A1)(ft) (C3)
Notes: Follow through from their answer to part (a). Accept 80.9° for using 12.9. Using the radian answer from part (a) leads to an impossible triangle, award (M1)(A1)(ft)(A0).
Question
The diagram shows a triangle \({\rm{ABC}}\). The size of angle \({\rm{C\hat AB}}\) is \(55^\circ\) and the length of \({\rm{AM}}\) is \(10\) m, where \({\rm{M}}\) is the midpoint of \({\rm{AB}}\). Triangle \({\rm{CMB}}\) is isosceles with \({\text{CM}} = {\text{MB}}\).
Write down the length of \({\rm{MB}}\).[1]
Find the size of angle \({\rm{C\hat MB}}\).[2]
Find the length of \({\rm{CB}}\).[3]
Answer/Explanation
Markscheme
\(10\) m (A1)(C1)
\({\rm{A\hat MC}} = 70^\circ \;\;\;\)OR\(\;\;\;{\rm{A\hat CM}} = 55^\circ \) (A1)
\({\rm{C\hat MB}} = 110^\circ \) (A1) (C2)
\({\text{C}}{{\text{B}}^2} = {10^2} + {10^2} – 2 \times 10 \times 10 \times \cos 110^\circ \) (M1)(A1)(ft)
Notes: Award (M1) for substitution into the cosine rule formula, (A1)(ft) for correct substitution. Follow through from their answer to part (b).
OR
\(\frac{{{\text{CB}}}}{{\sin 110^\circ }} = \frac{{10}}{{\sin 35^\circ }}\) (M1)(A1)(ft)
Notes: Award (M1) for substitution into the sine rule formula, (A1)(ft) for correct substitution. Follow through from their answer to part (b).
OR
\({\rm{A\hat CB}} = 90^\circ \) (A1)
\(\sin 55^\circ = \frac{{{\text{CB}}}}{{55}}\;\;\;\)OR\(\;\;\;\cos 35^\circ = \frac{{{\text{CB}}}}{{20}}\) (M1)
Note: Award (A1) for some indication that \({\rm{A\hat CB}} = 90^\circ \), (M1) for correct trigonometric equation.
OR
Perpendicular \({\rm{MN}}\) is drawn from \({\rm{M}}\) to \({\rm{CB}}\). (A1)
\(\frac{{\frac{1}{2}{\text{CB}}}}{{10}} = \cos 35^\circ \) (M1)
Note: Award (A1) for some indication of the perpendicular bisector of \({\rm{BC}}\), (M1) for correct trigonometric equation.
\({\text{CB}} = 16.4{\text{ (m)}}\;\;\;\left( {16.3830 \ldots {\text{ (m)}}} \right)\) (A1)(ft)(C3)
Notes: Where a candidate uses \({\rm{C\hat MB}} = 90^\circ \) and finds \({\text{CB}} = 14.1{\text{ (m)}}\) award, at most, (M1)(A1)(A0).
Where a candidate uses \({\rm{C\hat MB}} = 60^\circ \) and finds \({\text{CB}} = 10{\text{ (m)}}\) award, at most, (M1)(A1)(A0).
Question
In the following diagram, ABCD is the square base of a right pyramid with vertex V. The centre of the base is O. The diagonal of the base, AC, is 8 cm long. The sloping edges are 10 cm long.
Write down the length of \({\text{AO}}\).[1]
Find the size of the angle that the sloping edge \({\text{VA}}\) makes with the base of the pyramid.[2]
Hence, or otherwise, find the area of the triangle \({\text{CAV}}\).[3]
Answer/Explanation
Markscheme
\({\text{AO}} = 4{\text{ (cm)}}\) (A1) (C1)
\(\cos {\rm{O\hat AV}} = \frac{4}{{10}}\) (M1)
Note: Award (M1) for their correct trigonometric ratio.
OR
\(\cos {\rm{O\hat AV}} = \frac{{{{10}^2} + {8^2} – {{10}^2}}}{{2 \times 10 \times 8}}\;\;\;\)OR\(\;\;\;\frac{{{{10}^2} + {4^2} – {{(9.16515 \ldots )}^2}}}{{2 \times 10 \times 4}}\) (M1)
Note: Award (M1) for correct substitution into the cosine rule formula.
\({\rm{O\hat AV}} = 66.4^\circ \;\;\;(66.4218 \ldots )\) (A1)(ft) (C2)
Notes: Follow through from their answer to part (a).
\({\text{area}} = \frac{{8 \times 10 \times \sin (66.4218 \ldots ^\circ)}}{2}\;\;\;\)OR\(\;\;\;\frac{1}{2} \times 8 \times \sqrt {{{10}^2} – {4^{\text{2}}}} \)
OR\(\;\;\;\frac{1}{2} \times 10 \times 10 \times \sin (47.1563 \ldots ^\circ )\) (M1)(A1)(ft)
Notes: Award (M1) for substitution into the area formula, (A1)(ft) for correct substitutions. Follow through from their answer to part (b) and/or part (a).
\({\text{area}} = 36.7{\text{ c}}{{\text{m}}^2}\;\;\;(36.6606 \ldots {\text{ c}}{{\text{m}}^2})\) (A1)(ft) (C3)
Notes: Accept an answer of \(8\sqrt {21} {\text{ c}}{{\text{m}}^2}\) which is the exact answer.
Question
When Bermuda \({\text{(B)}}\), Puerto Rico \({\text{(P)}}\), and Miami \({\text{(M)}}\) are joined on a map using straight lines, a triangle is formed. This triangle is known as the Bermuda triangle.
According to the map, the distance \({\text{MB}}\) is \(1650\,{\text{km}}\), the distance \({\text{MP}}\) is \(1500\,{\text{km}}\) and angle \({\text{BMP}}\) is \(57^\circ \).
Calculate the distance from Bermuda to Puerto Rico, \({\text{BP}}\).[3]
Calculate the area of the Bermuda triangle.[3]
Answer/Explanation
Markscheme
\({\text{B}}{{\text{P}}^2} = {1650^2} + {1500^2} – 2 \times 1650 \times 1500\,\cos \,(57^\circ )\) (M1)(A1)
\(1510\,({\text{km}})\,\,\,\left( {1508.81…\,({\text{km}})} \right)\) (A1) (C3)
Notes: Award (M1) for substitution in the cosine rule formula, (A1) for correct substitution.
\(\frac{1}{2} \times 1650 \times 1500 \times \sin \,57^\circ \) (M1)(A1)
\( = 1\,040\,000\,({\text{k}}{{\text{m}}^2})\,\,\,\left( {1\,037\,854.82…\,({\text{k}}{{\text{m}}^2})} \right)\) (A1) (C3)
Note: Award (M1) for substitution in the area of triangle formula, (A1) for correct substitution.
Question
A triangular postage stamp, ABC, is shown in the diagram below, such that \({\text{AB}} = 5{\text{ cm}},{\rm{ B\hat AC}} = 34^\circ ,{\rm{ A\hat BC}} = 26^\circ \) and \({\rm{A\hat CB}} = 120^\circ \).
Find the length of BC.[3]
Find the area of the postage stamp.[3]
Answer/Explanation
Markscheme
\(\frac{{{\text{BC}}}}{{\sin 34^\circ }} = \frac{5}{{\sin 120^\circ }}\) (M1)(A1)
Note: Award (M1) for substituted sine rule formula, (A1) for correct substitutions.
\({\text{BC}} = 3.23{\text{ (cm) }}\left( {3.22850 \ldots {\text{ (cm)}}} \right)\) (A1) (C3)[3 marks]
\(\frac{1}{2}(5)(3.22850)\sin 26^\circ \) (M1)(A1)(ft)
Note: Award (M1) for substituted area of a triangle formula, (A1) for correct substitutions.
\( = 3.54{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}\left( {3.53820 \ldots {\text{ }}({\text{c}}{{\text{m}}^2})} \right)\) (A1)(ft) (C3)
Note: Follow through from part (a).[3 marks]
Question
Emily’s kite ABCD is hanging in a tree. The plane ABCDE is vertical.
Emily stands at point E at some distance from the tree, such that EAD is a straight line and angle BED = 7°. Emily knows BD = 1.2 metres and angle BDA = 53°, as shown in the diagram
T is a point at the base of the tree. ET is a horizontal line. The angle of elevation of A from E is 41°.
Find the length of EB.[3]
Write down the angle of elevation of B from E.[1]
Find the vertical height of B above the ground.[2]
Answer/Explanation
Markscheme
Units are required in parts (a) and (c).
\(\frac{{{\text{EB}}}}{{\sin 53{\rm{^\circ }}}} = \frac{{1.2}}{{\sin 7{\rm{^\circ }}}}\) (M1)(A1)
Note: Award (M1) for substitution into sine formula, (A1) for correct substitution.
\(({\text{EB}} = ){\text{ }}7.86{\text{ m}}\)\(\,\,\,\)OR\(\,\,\,\)\(786{\text{ cm }}(7.86385 \ldots {\text{ m}}\)\(\,\,\,\)OR\(\,\,\,\)\(786.385 \ldots {\text{ cm}})\) (A1) (C3)[3 marks]
34° (A1) (C1)[1 mark]
Units are required in parts (a) and (c).
\(\sin 34^\circ = \frac{{{\text{height}}}}{{7.86385 \ldots }}\) (M1)
Note: Award (M1) for correct substitution into a trigonometric ratio.
\(({\text{height}} = ){\text{ }}4.40{\text{ m}}\)\(\,\,\,\)OR\(\,\,\,\)\(440{\text{ cm }}(4.39741 \ldots {\text{ m}}\)\(\,\,\,\)OR\(\,\,\,\)\(439.741 \ldots {\text{ cm}})\) (A1)(ft) (C2)
Note: Accept “BT” used for height. Follow through from parts (a) and (b). Use of 7.86 gives an answer of 4.39525….[2 marks]
Question
Two fixed points, A and B, are 40 m apart on horizontal ground. Two straight ropes, AP and BP, are attached to the same point, P, on the base of a hot air balloon which is vertically above the line AB. The length of BP is 30 m and angle BAP is 48°.
Angle APB is acute.
On the diagram, draw and label with an x the angle of depression of B from P.[1]
Find the size of angle APB.[3]
Find the size of the angle of depression of B from P.[2]
Answer/Explanation
Markscheme
(A1) (C1)[1 mark]
\(\frac{{40}}{{{\text{sin APB}}}} = \frac{{30}}{{{\text{sin 48}}^\circ }}\) (M1)(A1)
Note: Award (M1) for substitution into sine rule, (A1) for correct substitution.
(angle APB =) 82.2° (82.2473…°) (A1) (C3)[3 marks]
180 − 48 − 82.2473… (M1)
49.8° (49.7526…°) (A1)(ft) (C2)
Note: Follow through from parts (a) and (b).[2 marks]
Question
A park in the form of a triangle, ABC, is shown in the following diagram. AB is 79 km and BC is 62 km. Angle A\(\mathop {\text{B}}\limits^ \wedge \)C is 52°.
Calculate the length of side AC in km.[3]
Calculate the area of the park.[3]
Answer/Explanation
Markscheme
(AC2 =) 622 + 792 − 2 × 62 × 79 × cos(52°) (M1)(A1)
Note: Award (M1) for substituting in the cosine rule formula, (A1) for correct substitution.
63.7 (63.6708…) (km) (A1) (C3)[3 marks]
\(\frac{1}{2}\) × 62 × 79 × sin(52°) (M1)(A1)
Note: Award (M1) for substituting in the area of triangle formula, (A1) for correct substitution.
1930 km2 (1929.83…km2) (A1) (C3)[3 marks]