References only, not question text. Open Official paper ↗ where Cambridge publishes the paper,
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school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Sine & cosine rules, ½ab sin C · Core · Paper 4 style · calculator · 3 marks
In triangle ABC, AB = 8.4 cm, AC = 6.1 cm and angle BAC = 52°.
(a)Calculate BC. Give your answer correct to 3 significant figures.[3]
Q4Sine & cosine rules, ½ab sin C · Standard · Paper 4 style · calculator · 3 marks
In triangle ABC, angle BAC = 47°, angle ABC = 68° and BC = 12.5 cm.
(a)Calculate AC. Give your answer correct to 3 significant figures.[3]
Show answer
(a) 15.8 cm
AC ÷ sin 68° = 12.5 ÷ sin 47° M2; AC = 15.847… = 15.8 A1
Q5Sine & cosine rules, ½ab sin C · Standard · Paper 4 style · calculator · 4 marks
A triangle has sides of length 7 cm, 9 cm and 12 cm.
(a)Calculate the size of the largest angle of the triangle. Give your answer correct to 1 decimal place.[4]
Show answer
(a) 96.4°
Largest angle is opposite the 12 cm side B1; cos θ = (7² + 9² − 12²) ÷ (2 × 7 × 9) M1; = −14/126 A1; 96.4 A1
Q6Sine & cosine rules, ½ab sin C · Standard · Paper 4 style · calculator · 3 marks
A triangle has area 54 cm². Two of its sides are 9 cm and x cm, and the angle between them is 40°.
(a)Find x. Give your answer correct to 3 significant figures.[3]
Show answer
(a) 18.7
½ × 9 × x × sin 40° = 54 M1; x = 108 ÷ (9 sin 40°) M1; 18.67 → 18.7 A1
Q7Sine & cosine rules, ½ab sin C · Challenge · Paper 4 style · calculator · 5 marks
In triangle ABC, AB = 9.2 cm, BC = 7.4 cm and angle BAC = 48°.
(a)Angle ACB is obtuse. Calculate angle ACB, correct to 1 decimal place.[4]
(b)Calculate angle ABC.[1]
Show answer
(a) 112.5°
sin C ÷ 9.2 = sin 48° ÷ 7.4 M1; sin C = 0.9239… A1; acute value 67.50° M1; obtuse angle 180 − 67.50 = 112.5° A1
(b) 19.5°
180 − 48 − 112.5 B1
Q8Sine & cosine rules, ½ab sin C · Challenge · Paper 4 style · calculator · 9 marks
A field is in the shape of a quadrilateral ABCD. AB = 120 m, BC = 95 m and angle ABC = 108°. The diagonal AC divides the field into two triangles. Angle CAD = 40° and angle ACD = 65°. Keep full calculator values for later parts.
(a)Calculate AC.[3]
(b)Calculate CD.[3]
(c)Calculate the area of the field ABCD, correct to 3 significant figures.[3]
Q10Sine & cosine rules, ½ab sin C · Challenge · Paper 2 style · no calculator · 5 marks
Do not use a calculator in this question.
(a)A triangle has sides 6 cm and 10 cm with an angle of 30° between them. Find its area.[2]
(b)Another triangle has sides 8 cm and 6 cm and area 12√3 cm². Find the two possible sizes of the angle between these sides.[3]
Show answer
(a) 15 cm²
½ × 6 × 10 × sin 30° M1; sin 30° = ½, 15 A1
(b) 60° or 120°
½ × 8 × 6 × sin θ = 12√3 M1; sin θ = √3/2 A1; 60° and 120° A1
Q11Right-angled trigonometry · Core · Paper 3 style · calculator · 2 marks
A ladder 6.5 m long leans against a vertical wall on horizontal ground. The ladder makes an angle of 72° with the ground.
(a)Calculate how far up the wall the ladder reaches. Give your answer correct to 3 significant figures.[2]
Show answer
(a) 6.18 m
6.5 × sin 72° M1; 6.18 A1
Q12Right-angled trigonometry · Core · Paper 3 style · calculator · 2 marks
Triangle ABC has a right angle at B. AB = 11.2 cm and BC = 7.3 cm.
(a)Calculate angle BAC. Give your answer correct to 1 decimal place.[2]
Show answer
(a) 33.1°
tan BAC = 7.3 ÷ 11.2 M1; 33.1 A1
Q13Right-angled trigonometry · Standard · Paper 4 style · calculator · 3 marks
Usman stands on level ground 45 m from the foot of a vertical minaret. His eyes are 1.6 m above the ground. The angle of elevation of the top of the minaret from his eyes is 38°.
(a)Calculate the height of the minaret. Give your answer correct to 3 significant figures.[3]
Show answer
(a) 36.8 m
45 × tan 38° = 35.16 M1 A1; + 1.6 = 36.8 A1
Q14Right-angled trigonometry · Standard · Paper 3 style · calculator · 5 marks
Ayesha walks 5 km due north and then 3.2 km due east.
(a)Calculate how far she is from her starting point. Give your answer correct to 3 significant figures.[2]
(b)Calculate the bearing of her final position from her starting point, correct to the nearest 0.1°.[3]
Q15Right-angled trigonometry · Standard · Paper 3 style · calculator · 4 marks
A wheelchair ramp rises 0.75 m. The ramp makes an angle of 4.8° with the horizontal.
(a)Calculate the length of the sloping surface of the ramp. Give your answer correct to 3 significant figures.[2]
(b)Calculate the horizontal distance covered by the ramp. Give your answer correct to 3 significant figures.[2]
Show answer
(a) 8.96 m
0.75 ÷ sin 4.8° M1; 8.96 A1
(b) 8.93 m
0.75 ÷ tan 4.8° M1; 8.93 A1
Q16Right-angled trigonometry · Challenge · Paper 4 style · calculator · 4 marks
From the top of a vertical cliff 80 m above sea level, the angle of depression of a boat is 32°. The boat sails directly away from the cliff until the angle of depression is 18°.
(a)Calculate the distance the boat moves. Give your answer correct to 3 significant figures.[4]
Show answer
(a) 118 m
Distance at first = 80 ÷ tan 32° = 128.0 M1; later = 80 ÷ tan 18° = 246.2 M1; 246.2 − 128.0 M1; 118 A1
Q173D Pythagoras & trigonometry · Core · Paper 4 style · calculator · 5 marks
ABCDEFGH is a cuboid. The base ABCD has AB = 8 cm and BC = 6 cm. The vertical edges AE, BF, CG and DH are 5 cm. G is directly above C.
(a)Calculate the length of the diagonal AG. Give your answer correct to 3 significant figures.[2]
(b)Calculate the angle between AG and the base ABCD.[3]
Show answer
(a) 11.2 cm
√(8² + 6² + 5²) = √125 M1; 11.2 A1
(b) 26.6°
AC = √(8² + 6²) = 10 M1; tan θ = 5 ÷ 10 M1; 26.6 A1
Q183D Pythagoras & trigonometry · Standard · Paper 4 style · calculator · 5 marks
VABCD is a pyramid with a square base ABCD of side 10 cm. The vertex V is vertically above the centre of the base, and the vertical height is 12 cm.
(a)Calculate the length of a sloping edge, VA. Give your answer correct to 3 significant figures.[3]
(b)Calculate the angle between VA and the base.[2]
Show answer
(a) 13.9 cm
Half-diagonal of base = √50 = 7.071 M1; VA = √(50 + 12²) = √194 M1; 13.9 A1
(b) 59.5°
tan θ = 12 ÷ √50 M1; 59.5 A1
Q193D Pythagoras & trigonometry · Standard · Paper 4 style · calculator · 5 marks
A school field is a horizontal rectangle ABCD with AB = 40 m and BC = 30 m. A vertical flagpole 18 m tall stands at corner A.
(a)Calculate the distance AC.[1]
(b)Calculate the angle of elevation of the top of the flagpole from C.[2]
(c)Calculate the angle of elevation of the top of the flagpole from B.[2]
Show answer
(a) 50 m
√(40² + 30²) B1
(b) 19.8°
tan θ = 18 ÷ 50 M1; 19.8 A1
(c) 24.2°
tan θ = 18 ÷ 40 M1; 24.2 A1
Q203D Pythagoras & trigonometry · Challenge · Paper 4 style · calculator · 3 marks
A cube has edge length a cm. A space diagonal joins two opposite vertices of the cube, passing through its centre.
(a)Calculate the angle between a space diagonal of the cube and one of its faces. Give your answer correct to 1 decimal place.[3]
Show answer
(a) 35.3°
Face diagonal = a√2 M1; tan θ = a ÷ (a√2) = 1/√2 M1; 35.3 A1
Q213D Pythagoras & trigonometry · Challenge · Paper 4 style · calculator · 5 marks
A ramp has a horizontal rectangular base ABCD with AB = 12 m and BC = 7 m. BCFE is a vertical rectangle with BE = CF = 4 m, so E is directly above B and F is directly above C. ADFE is the sloping surface.
(a)Calculate the length AF.[3]
(b)Calculate the angle between AF and the base ABCD.[2]