O Level Mathematics D (4024)
Trigonometry Banker
30 questions · about 5% of the marks we indexed · examined in 9 of 9 sessions.
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 · Core · Paper 2 style · calculator · 5 marksIn triangle ABC, AB = 8 cm, AC = 11 cm and angle BAC = 52°. Give your answers correct to 3 significant figures.
- (a) Calculate BC. [3]
- (b) Calculate the area of triangle ABC. [2]
Show answer
- (a) 8.75 cm
BC² = 8² + 11² − 2 × 8 × 11 × cos 52° M1; = 76.64… A1; BC = 8.75 A1
- (b) 34.7 cm²
½ × 8 × 11 × sin 52° M1; 34.7 A1
In triangle PQR, QR = 9.5 cm, angle QPR = 41° and angle PQR = 73°.
- (a) Calculate PR, correct to 3 significant figures. [3]
- (b) Find angle PRQ. [1]
Show answer
- (a) 13.8 cm
PR/sin 73° = 9.5/sin 41° M2 (M1 for correct pairing); A1
- (b) 66°
B1
A triangle has sides of length 7 cm, 9 cm and 12 cm.
- (a) Calculate the largest angle of the triangle, correct to 1 decimal place. [3]
- (b) Calculate the area of the triangle, correct to 3 significant figures. [2]
Show answer
- (a) 96.4°
Largest angle is opposite 12 cm B1; cos θ = (49 + 81 − 144)/(2 × 7 × 9) M1; A1
- (b) 31.3 cm²
½ × 7 × 9 × sin θ M1 A1
A triangular plot of land has two sides of 25 m and 32 m with an angle of 118° between them.
- (a) Calculate the area of the plot, correct to 3 significant figures. [2]
- (b) Grass seed costs Rs 180 per m². Using your answer to part (a), find the cost of seeding the whole plot. [1]
- (c) Calculate the length of the third side, correct to 3 significant figures. [3]
Show answer
- (a) 353 m²
½ × 25 × 32 × sin 118° M1 A1
- (b) Rs 63540
353 × 180 B1
- (c) 49.0 m
c² = 25² + 32² − 2 × 25 × 32 × cos 118° M1; = 2400.1… A1; A1
In triangle XYZ, XY = 5.2 cm, XZ = 7.8 cm and angle YXZ = 63°.
- (a) Calculate YZ, correct to 3 significant figures. [3]
- (b) Calculate angle XYZ, correct to 1 decimal place. [2]
Show answer
- (a) 7.15 cm
YZ² = 5.2² + 7.8² − 2 × 5.2 × 7.8 × cos 63° M1; A1; A1
- (b) 76.6°
Cosine rule (or sine rule, taking care that the angle is obtuse or acute) M1 A1
In triangle ABC, BC = 14 cm, angle ABC = 48° and angle ACB = 67°. Give lengths correct to 3 significant figures.
- (a) Find angle BAC. [1]
- (b) Calculate AB. [2]
- (c) Calculate AC. [2]
Show answer
- (a) 65°
180 − 48 − 67 B1
- (b) 14.2 cm
AB/sin 67° = 14/sin 65° M1; 14.2 A1
- (c) 11.5 cm
AC = 14 sin 48°/sin 65° M1; 11.5 A1
Two ships leave a port at the same time. Ship A sails 14 km on a bearing of 035°. Ship B sails 20 km on a bearing of 290°.
- (a) Show that the angle between the directions of the two ships is 105°. [1]
- (b) Calculate the distance between the ships, correct to 3 significant figures. [3]
- (c) Calculate the area of the triangle formed by the port and the two ships, correct to 3 significant figures. [2]
Show answer
- (a) (360 − 290) + 35 = 105°
B1
- (b) 27.2 km
d² = 14² + 20² − 2 × 14 × 20 × cos 105° M1; = 740.9… A1; A1
- (c) 135 km²
½ × 14 × 20 × sin 105° M1 A1
In triangle ABC, CA = 9 cm and CB = 12 cm. The area of the triangle is 40 cm².
- (a) Find angle C, given that it is acute. Give your answer correct to 1 decimal place. [3]
- (b) Write down the other possible value of angle C if it is not acute. [1]
Show answer
- (a) 47.8°
½ × 9 × 12 × sin C = 40 M1; sin C = 80/108 A1; A1
- (b) 132.2°
180 − answer to (a) B1
A, B and the foot F of a vertical tower lie in a straight line on level ground, in that order. AB = 50 m. T is the top of the tower. The angle of elevation of T from A is 28° and from B is 43°.
- (a) Find angle ATB. [1]
- (b) Use the sine rule in triangle ABT to calculate BT, correct to 3 significant figures. [2]
- (c) Calculate the height of the tower, correct to 3 significant figures. [2]
Show answer
- (a) 15°
Exterior angle of triangle ABT: 43 − 28 B1
- (b) 90.7 m
BT/sin 28° = 50/sin 15° M1 A1
- (c) 61.9 m
BT sin 43° M1 A1
A field PQRS is a quadrilateral. PQ = 7 m, QR = 9 m and angle PQR = 110°. The diagonal PR divides the field into two triangles. Give answers correct to 3 significant figures, angles to 1 decimal place.
- (a) Calculate PR. [3]
- (b) PS = 12 m and RS = 8 m, and S and Q are on opposite sides of PR. Calculate angle PSR. [3]
- (c) Calculate the area of the field PQRS. [2]
Show answer
- (a) 13.2 m
PR² = 7² + 9² − 2 × 7 × 9 × cos 110° M1; = 173.09… A1; 13.2 A1
- (b) 79.5°
cos S = (12² + 8² − PR²)/(2 × 12 × 8) M1; = 0.1818… A1; 79.5° A1
- (c) 76.8 m²
½ × 7 × 9 × sin 110° + ½ × 12 × 8 × sin 79.5° M1; 76.8 A1
In triangle ABC, AB = 10 cm, BC = 7 cm and angle BAC = 40°. Angle ACB is obtuse.
- (a) Show that sin ACB = 0.918, correct to 3 decimal places. [2]
- (b) Find angle ACB, correct to 1 decimal place. [2]
- (c) Calculate AC, correct to 3 significant figures. [2]
Show answer
- (a) sin C = 10 sin 40°/7 = 0.9183
sin C/10 = sin 40°/7 M1; A1
- (b) 113.3°
Acute value 66.7° M1; obtuse angle 180 − 66.7 A1
- (c) 4.89 cm
Angle ABC = 180 − 40 − 113.3 = 26.7° M1; AC = 7 sin 26.7°/sin 40° A1
ABCD is a quadrilateral. AB = 6 cm, BC = 8 cm, angle ABC = 100°, CD = 7 cm and angle ACD = 52°.
- (a) Calculate AC, correct to 3 significant figures. [3]
- (b) Calculate AD, correct to 3 significant figures. [3]
- (c) Calculate the area of quadrilateral ABCD, correct to 3 significant figures. [2]
Show answer
- (a) 10.8 cm
AC² = 36 + 64 − 96 cos 100° M1; = 116.67… A1; A1
- (b) 8.52 cm
AD² = AC² + 7² − 2 × AC × 7 × cos 52° M1; A1; A1
- (c) 53.4 cm²
½ × 6 × 8 × sin 100° + ½ × AC × 7 × sin 52° M1 A1
A triangle has sides x cm, (x + 1) cm and (x + 2) cm. The largest angle of the triangle is 120°.
- (a) Use the cosine rule to show that 2x² − x − 3 = 0. [cos 120° = −½] [3]
- (b) Solve the equation and write down the lengths of the three sides. [2]
Show answer
- (a) (x + 2)² = x² + (x + 1)² + x(x + 1), which simplifies to 2x² − x − 3 = 0
Cosine rule with the longest side opposite 120° M1; −2x(x + 1)(−½) = + x(x + 1) M1; simplify A1
- (b) x = 1.5; sides 1.5 cm, 2.5 cm, 3.5 cm
(2x − 3)(x + 1) = 0 M1; x = 1.5 (x > 0) A1
- (a) A right-angled triangle has shorter sides 9 cm and 12 cm. Find the length of the hypotenuse. [2]
- (b) Another right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. Find the length of the third side. [2]
Show answer
- (a) 15 cm
√(81 + 144) = √225 M1; 15 A1
- (b) 12 cm
√(169 − 25) = √144 M1; 12 A1
- (a) A rectangular television screen is 48 cm wide and 36 cm high. Find the length of its diagonal. [2]
- (b) An isosceles triangle has two sides of 13 cm and a base of 10 cm. Find its height and its area. [3]
Show answer
- (a) 60 cm
√(2304 + 1296) M1 A1
- (b) Height 12 cm, area 60 cm²
Half base 5 M1; √(169 − 25) = 12 A1; ½ × 10 × 12 = 60 B1
A ladder 6.5 m long leans against a vertical wall. The foot of the ladder is on horizontal ground, 1.8 m from the wall.
- (a) How high up the wall does the ladder reach? Give your answer correct to 3 significant figures. [2]
- (b) The foot of the ladder slides 0.7 m further from the wall. How far does the top of the ladder slide down the wall? Give your answer correct to 2 significant figures. [3]
Show answer
- (a) 6.25 m
√(6.5² − 1.8²) = √39.01 M1; 6.25 A1
- (b) 0.25 m
New base 2.5 m B1; √(42.25 − 6.25) = 6 M1; 6.2458 − 6 = 0.25 A1
A is the point (−3, 4) and B is the point (5, −2).
- (a) Find the length of AB. [2]
- (b) C is the point (k, 4), where k ≠ −3, and BC = 10. Find k. [3]
- (c) Find the area of triangle ABC. [2]
Show answer
- (a) 10 units
√(8² + 6²) M1; 10 A1
- (b) k = 13
(k − 5)² + 6² = 100 M1; k − 5 = ±8 M1; k = 13 (k = −3 gives A) A1
- (c) 48 square units
Base AC = 16, height 6 M1; 48 A1
A right-angled triangle has sides x cm, (x + 7) cm and (x + 8) cm.
- (a) Show that x² − 2x − 15 = 0. [2]
- (b) Find x and the perimeter of the triangle. [3]
Show answer
- (a) x² + (x + 7)² = (x + 8)² simplifies to x² − 2x − 15 = 0
Pythagoras with hypotenuse x + 8 M1; expand and simplify A1
- (b) x = 5; perimeter 30 cm
(x − 5)(x + 3) = 0 M1; x = 5 A1; 5 + 12 + 13 B1
Give your answers correct to 3 significant figures.
- (a) From a point on level ground 40 m from the base of a tower, the angle of elevation of the top is 32°. Calculate the height of the tower. [2]
- (b) From the top of a cliff 75 m high, the angle of depression of a boat is 18°. Calculate the horizontal distance from the foot of the cliff to the boat. [2]
Show answer
- (a) 25.0 m
40 tan 32° M1; 25.0 A1
- (b) 231 m
75 ÷ tan 18° M1; 231 A1
A cuboid has a horizontal rectangular base EFGH with EF = 8 cm and FG = 6 cm. The top face is ABCD, with A above E, B above F, C above G and D above H. The height of the cuboid is 5 cm. Give lengths correct to 3 significant figures and angles to 1 decimal place.
- (a) Calculate the length of the diagonal EG of the base. [2]
- (b) Calculate the length of the diagonal EC. [2]
- (c) Calculate the angle between EC and the base EFGH. [2]
Show answer
- (a) 10 cm
√(8² + 6²) M1; 10 A1
- (b) 11.2 cm
√(10² + 5²) = √125 M1; 11.2 A1
- (c) 26.6°
tan⁻¹(5/10) M1; 26.6° A1
VABCD is a pyramid with a horizontal square base ABCD of side 10 cm. M is the centre of the base and V is vertically above M. Each sloping edge (VA, VB, VC, VD) is 13 cm. Give lengths correct to 3 significant figures and angles to 1 decimal place.
- (a) Calculate the height of the pyramid. [3]
- (b) Calculate the angle between the edge VA and the base. [2]
- (c) N is the midpoint of BC. Calculate the angle between the face VBC and the base. [2]
Show answer
- (a) 10.9 cm
AM = ½√200 = 7.071 M1; h = √(13² − 50) = √119 M1; 10.9 A1
- (b) 57.0°
cos⁻¹(7.071/13) M1; 57.0° A1
- (c) 65.4°
MN = 5, tan⁻¹(√119/5) M1; 65.4° A1
A vertical mast BT of height h m stands at B on horizontal ground. Point A is due south of B and point C is due east of B. The angle of elevation of T from A is 24° and from C is 31°. AC = 150 m.
- (a) Write AB and CB in terms of h. [2]
- (b) Explain why angle ABC = 90° and hence find h, correct to 3 significant figures. [4]
- (c) Find the bearing of C from A, correct to the nearest degree. [2]
Show answer
- (a) AB = h/tan 24°, CB = h/tan 31°
B1 B1
- (b) A is due south and C is due east of B, so angle ABC = 90°; h = 53.7 m
Right angle B1; (h/tan 24°)² + (h/tan 31°)² = 150² M1; h² (5.04… + 2.77…) = 22500 M1; A1
- (c) 037°
tan(angle at A) = CB/AB M1; bearing measured from north A1
Give lengths correct to 3 significant figures and angles to 1 decimal place.
- (a) Triangle PQR has a right angle at Q. PR = 12 cm and angle QPR = 35°. Calculate QR and PQ. [3]
- (b) Triangle XYZ has a right angle at Y. XY = 11 cm and YZ = 7 cm. Calculate angle YXZ. [2]
Show answer
- (a) QR = 6.88 cm, PQ = 9.83 cm
12 sin 35° M1 A1; 12 cos 35° = 9.83 A1
- (b) 32.5°
tan⁻¹(7/11) M1; 32.5° A1
A straight wheelchair ramp is 4.5 m long and rises a vertical height of 0.6 m.
- (a) Calculate the angle the ramp makes with the horizontal. Give your answer correct to 1 decimal place. [2]
- (b) Calculate the horizontal distance covered by the ramp. Give your answer correct to 3 significant figures. [2]
- (c) Safety rules say the angle must be less than 8°. Does the ramp meet the rule? [1]
Show answer
- (a) 7.7°
sin⁻¹(0.6/4.5) M1; 7.7° A1
- (b) 4.46 m
√(4.5² − 0.6²) or 4.5 cos 7.66° M1; 4.46 A1
- (c) Yes, 7.7° < 8°
B1 ft
Ali holds a kite string 1.2 m above level ground. The string is straight, 60 m long, and makes an angle of 52° with the horizontal.
- (a) Calculate the height of the kite above the ground, correct to 3 significant figures. [2]
- (b) Calculate the horizontal distance from Ali's hand to the point directly below the kite, correct to 3 significant figures. [2]
- (c) Ali lets out more string, so the string is 75 m long and the kite is 62 m above the ground. Find the new angle of the string with the horizontal, correct to 1 decimal place. [2]
Show answer
- (a) 48.5 m
60 sin 52° M1; + 1.2 A1
- (b) 36.9 m
60 cos 52° M1 A1
- (c) 54.2°
sin θ = 60.8/75 M1 A1
AB is a vertical pole of height 9 m standing at B on horizontal ground. B, D and C lie on a straight line on the ground, with D between B and C. Angle ADB = 50° and angle ACB = 28°. Give your answers correct to 3 significant figures.
- (a) Calculate BD. [2]
- (b) Calculate DC. [3]
- (c) Calculate AD. [2]
Show answer
- (a) 7.55 m
9 ÷ tan 50° M1; 7.55 A1
- (b) 9.37 m
BC = 9 ÷ tan 28° = 16.93 M1 A1; DC = 16.93 − 7.55 = 9.37 A1
- (c) 11.7 m
9 ÷ sin 50° M1; 11.7 A1
- (a) The bearing of B from A is 065°. Find the bearing of A from B. [1]
- (b) The bearing of D from C is 300°. Find the bearing of C from D. [1]
- (c) Write down the three-figure bearing of south-west. [1]
Show answer
- (a) 245°
65 + 180 B1
- (b) 120°
300 − 180 B1
- (c) 225°
B1
A ship sails 12 km from a port on a bearing of 040°. It then sails 9 km on a bearing of 130°.
- (a) Explain why the angle between the two legs of the journey is 90°. [1]
- (b) Calculate the distance of the ship from its starting point. [2]
- (c) Calculate the bearing of the ship from its starting point, correct to 1 decimal place. [2]
- (d) Find the bearing of the starting point from the ship. [1]
Show answer
- (a) Bearing back to the start from the turning point is 220°; 220° − 130° = 90°.
B1
- (b) 15 km
√(12² + 9²) M1; 15 A1
- (c) 076.9°
tan⁻¹(9/12) = 36.87° M1; 40 + 36.87 = 076.9° A1
- (d) 256.9°
76.9 + 180 B1 ft
A drone flies 8 km from A to B on a bearing of 070°. It then flies 11 km from B to C on a bearing of 185°.
- (a) Show that angle ABC = 65°. [1]
- (b) Calculate AC, correct to 3 significant figures. [3]
- (c) Calculate the bearing of C from A, correct to 1 decimal place. [3]
Show answer
- (a) Bearing of A from B is 250°; 250° − 185° = 65°
B1
- (b) 10.5 km
AC² = 8² + 11² − 2 × 8 × 11 × cos 65° M1; = 110.6… A1; 10.5 A1
- (c) 141.4°
sin BAC = 11 sin 65°/10.52 M1; angle BAC = 71.4° A1; 70 + 71.4 = 141.4° A1
Do not use a calculator in this question.
- (a) Find the exact value of 2 sin 30° + tan 45°. [2]
- (b) Find the exact value of cos 30° × tan 60°. [2]
- (c) Solve sin x = ½ for 0° ≤ x ≤ 180°. [2]
Show answer
- (a) 2
2 × ½ + 1 M1; 2 A1
- (b) 3/2
(√3/2) × √3 M1; 3/2 A1
- (c) 30°, 150°
B1 B1
Give your answers correct to 1 decimal place.
- (a) Solve 3 sin x = 2 for 0° ≤ x ≤ 360°. [3]
- (b) Solve cos x = −0.4 for 0° ≤ x ≤ 360°. [3]
Show answer
- (a) 41.8°, 138.2°
sin x = 2/3 M1; 41.8° A1; 180 − 41.8 = 138.2° A1
- (b) 113.6°, 246.4°
cos⁻¹(−0.4) M1; 113.6° A1; 360 − 113.6 = 246.4° A1
Do not use a calculator in this question.
- (a) Given that cos 70° = 0.342, write down the three other values of x between 0° and 360° for which cos x = 0.342 or cos x = −0.342, and say which equation each solves. [3]
- (b) Solve tan x = √3 for 0° ≤ x ≤ 360°. [2]
- (c) For y = 2 sin x + 1, 0° ≤ x ≤ 360°, write down the maximum and minimum values of y and the values of x where they occur. [2]
Show answer
- (a) cos 290° = 0.342; cos 110° = −0.342; cos 250° = −0.342
Use symmetry of y = cos x: 360 − 70 B1; 180 − 70 B1; 180 + 70 B1
- (b) 60°, 240°
B1; +180° B1
- (c) Max 3 at x = 90°; min −1 at x = 270°
B1 B1
