Trigonometry Practice Questions – Extended Level
Extended-level trigonometry questions for IGCSE 0580 Papers 2 and 4. Covers sine rule, cosine rule, 3D trig and more.
Questions
In triangle ABC, AB = 8 cm, angle A = 50° and angle B = 70°. Use the sine rule to find BC.
In triangle PQR, PQ = 10 cm, QR = 7 cm and angle Q = 120°. Find PR using the cosine rule.
Find the area of triangle XYZ where XY = 9 cm, XZ = 12 cm and angle X = 40°.
A vertical pole AB stands on horizontal ground. From a point C on the ground, 20 m from B, the angle of elevation of A is 35°. From a point D on the ground, the angle of elevation of A is 25°. C, B and D are in a straight line. Find CD.
A ship sails from port P on a bearing of 040° for 15 km to point Q, then on a bearing of 130° for 20 km to point R. Find the distance PR and the bearing of R from P.
Solve sin x = 0.6 for 0° ≤ x ≤ 360°.
A cuboid has length 12 cm, width 5 cm and height 8 cm. Find the angle between the space diagonal AG and the base ABCD.
Sketch the graph of y = 2 sin x for 0° ≤ x ≤ 360°. State the amplitude and period.
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Question 1 Solution
Step 1: Angle C = 180° − 50° − 70° = 60°
Step 2: BC/sin A = AB/sin C
Step 3: BC/sin 50° = 8/sin 60°
Step 4: BC = 8 × sin 50° / sin 60° = 8 × 0.7660 / 0.8660 = 7.08 cm
Answer: 7.08 cm
Question 2 Solution
Step 1: PR² = PQ² + QR² − 2(PQ)(QR)cos Q
Step 2: PR² = 100 + 49 − 2(10)(7)cos 120°
Step 3: cos 120° = −0.5, so PR² = 149 − 140(−0.5) = 149 + 70 = 219
Step 4: PR = √219 = 14.8 cm
Answer: 14.8 cm
Question 3 Solution
Step 1: Area = ½ab sin C = ½ × 9 × 12 × sin 40°
Step 2: Area = 54 × 0.6428 = 34.7 cm²
Answer: 34.7 cm²
Question 4 Solution
Step 1: Height AB = 20 tan 35° = 20 × 0.7002 = 14.00 m
Step 2: BD = AB / tan 25° = 14.00 / 0.4663 = 30.02 m
Step 3: CD = BD − BC = 30.02 − 20 = 10.02 m
Answer: 10.0 m
Question 5 Solution
Step 1: Angle PQR = 180° − 130° + 40° = 90° (angles on bearings)
Step 2: Since angle PQR = 90°, use Pythagoras: PR² = 15² + 20² = 225 + 400 = 625
Step 3: PR = 25 km
Step 4: tan(angle QPR) = 20/15, angle QPR = 53.1°
Step 5: Bearing of R from P = 040° + 53.1° = 093.1°
Answer: PR = 25 km, bearing 093.1°
Question 6 Solution
Step 1: sin⁻¹(0.6) = 36.87°
Step 2: sin is positive in 1st and 2nd quadrants
Step 3: x = 36.9° or x = 180° − 36.87° = 143.1°
Answer: x = 36.9° or x = 143.1°
Question 7 Solution
Step 1: Base diagonal AC = √(12² + 5²) = √(144+25) = √169 = 13 cm
Step 2: Space diagonal AG goes from A to opposite corner G, height = 8 cm
Step 3: tan θ = 8/13
Step 4: angle = tan⁻¹(8/13) = 31.6°
Answer: 31.6°
Question 8 Solution
Step 1: The graph is a sine curve stretched vertically by factor 2
Step 2: Maximum value = 2, minimum value = −2
Step 3: Amplitude = 2, period = 360°
Step 4: Key points: (0,0), (90,2), (180,0), (270,−2), (360,0)
Answer: Amplitude = 2, Period = 360°
Frequently Asked Questions
When do I use sine rule vs cosine rule?
Use the sine rule when you have a pair of opposite side and angle (AAS or ASA). Use the cosine rule when you have SAS (two sides and included angle) or SSS (three sides).
How do I handle 3D trigonometry problems?
Break the 3D problem into 2D right-angled triangles. Usually you find a length in the base first, then use it with the height to find angles or the space diagonal.
What are common bearing mistakes?
Always measure bearings clockwise from North using three figures (e.g. 045° not 45°). Draw a North line at each point. The bearing FROM A to B is different from the bearing FROM B to A.
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