This set has eleven original questions, ordered from easier to harder, covering all five lessons in non-right triangles and bearings. Questions 1 to 4 are warm-ups, 5 to 8 need a full method, and 9 to 11 mix skills and include a second-triangle check.
Use a calculator in degree mode. Sketch every diagram first, and give answers to 3 significant figures, and angles to 1 decimal place unless a question says otherwise. The triangle and bearings reasoning board and the mistake log and retest queue are helpful companions while you work.
Questions
1. Which rule would you use first in each case: (a) sides 7, 9 and 12, find an angle; (b) angle A = 40°, angle B = 65°, a = 8, find b; (c) two sides and the angle between them, find the third side?
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(a) No angle is known, all sides are known: cosine rule. (b) Side a with its opposite angle A, plus angle B: sine rule. (c) Two sides and the included angle: cosine rule.
2. (a) Write 9° as a three-figure bearing. (b) The bearing of E from F is 072°. Find the bearing of F from E. (c) The bearing of G from H is 305°. Find the bearing of H from G.
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(a) 009°. (b) 72° is below 180°, so add: 72° + 180° = 252°. (c) 305° is above 180°, so subtract: 305° − 180° = 125°.
3. A man walks 15 km on a bearing of 250°. How far west and how far south of his start is he?
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The angle from the south line is 250° − 180° = 70°. West = 15 × sin 70° = 15 × 0.9397 = 14.10 km. South = 15 × cos 70° = 15 × 0.3420 = 5.13 km. Check: 14.10² + 5.13² = 198.8 + 26.3 = 225.1, near 15² = 225. The man is 14.1 km west and 5.13 km south.
4. In triangle ABC, angle A = 64°, angle B = 48° and BC = 11.5 cm. Find AC.
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BC is opposite A, and AC is opposite B. Sine rule: AC / sin 48° = 11.5 / sin 64°. AC = 11.5 × 0.7431 ÷ 0.8988 = 8.546 ÷ 0.8988 = 9.508. So AC = 9.51 cm.
5. In triangle PQR, PQ = 8.3 cm, PR = 5.6 cm and angle P = 71°. Find QR.
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Cosine rule: QR² = 8.3² + 5.6² − 2 × 8.3 × 5.6 × cos 71° = 68.89 + 31.36 − 92.96 × 0.3256 = 100.25 − 30.26 = 69.99. So QR = 8.366, giving QR = 8.37 cm. Multiply before subtracting.
6. A triangle has sides 7 cm, 9 cm and 12 cm. Find its largest angle.
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The largest angle faces the longest side, 12 cm. cos θ = (7² + 9² − 12²) ÷ (2 × 7 × 9) = (49 + 81 − 144) ÷ 126 = −14 ÷ 126 = −0.1111. So θ = 96.4°. The negative cosine tells you the angle is obtuse.
7. Two sides of a triangle are 14 cm and 9 cm, and the angle between them is 48°. Find the area.
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Area = ½ × 14 × 9 × sin 48° = 63 × 0.7431 = 46.82. So the area is 46.8 cm².
8. A ship sails 12 km on a bearing of 075° from A to B, then 9 km on a bearing of 160° to C. Find (a) angle ABC, (b) the distance AC, (c) the bearing of C from A.
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(a) The bearing of A from B is 075° + 180° = 255°. Angle ABC = 255° − 160° = 95°.
(b) Cosine rule: AC² = 12² + 9² − 2 × 12 × 9 × cos 95° = 225 + 216 × 0.08716 = 225 + 18.83 = 243.8. AC = 15.61, so 15.6 km.
(c) Sine rule: sin A = 9 × sin 95° ÷ 15.61 = 9 × 0.9962 ÷ 15.61 = 0.5742, so angle BAC = 35.0°. Bearing = 075° + 35.0° = 110°.
Check: east = 12 sin 75° + 9 sin 160° = 11.59 + 3.08 = 14.67; north = 12 cos 75° + 9 cos 160° = 3.11 − 8.46 = −5.35. The point is east and slightly south of A, and the angle from north is 110.0°.
9. A triangle has area 27 cm². Two of its sides are 10 cm and 6.5 cm. Find the two possible sizes of the angle between them.
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27 = ½ × 10 × 6.5 × sin C = 32.5 sin C. So sin C = 27 ÷ 32.5 = 0.8308. The acute angle is 56.2°. The obtuse angle is 180° − 56.2° = 123.8°. Answer: 56.2° or 123.8°. Both have the same sine, so both give an area of 27 cm².
10. In triangle ABC, angle A = 28°, BC = 7 cm and AB = 12 cm. (a) Explain why there may be two triangles. (b) Find the two possible sizes of angle C.
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(a) c sin A = 12 × 0.4695 = 5.63. Since 5.63 < 7 < 12, side a can meet the third side in two places, so two triangles are possible.
(b) sin C = 12 × sin 28° ÷ 7 = 0.8048. C = 53.6° or 180° − 53.6° = 126.4°. Check the angle sums: 28° + 53.6° = 81.6°, and 28° + 126.4° = 154.4°. Both are below 180°, so both triangles exist.
11. A triangular field PQR has PQ = 120 m, QR = 150 m and angle Q = 62°. Find (a) the area of the field, (b) the length PR.
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(a) Area = ½ × 120 × 150 × sin 62° = 9000 × 0.8829 = 7946.5. So 7950 m² (3 s.f.).
(b) Cosine rule: PR² = 120² + 150² − 2 × 120 × 150 × cos 62° = 36 900 − 36 000 × 0.4695 = 36 900 − 16 901 = 19 999. So PR = 141.4, giving 141 m.
If you got these wrong
Match each error to the lesson that repairs it.
| What went wrong | Go back to |
|---|---|
| Not sure which rule to start with (questions 1, 4, 5, 6) | Choose a sine or cosine relationship |
| Bearing written without three figures, or back bearing wrong (question 2, 3) | Use bearings with north lines consistently |
| Wrong angle inside a two-leg journey (question 8) | Resolve a two-stage journey diagram |
| Area angle not between the two sides, or only one angle given (questions 7, 9, 11) | Find an area from two sides and an included angle |
| Only one triangle found when two fit (questions 9, 10) | Explain an ambiguous diagram before calculating |
The non-calculator working trainer helps with the exact values that appear in these questions, and the mistake log and retest queue keeps your repeat errors in one place.
If the same error keeps coming back, it is worth having a teacher watch you work one question from start to finish. That is what online one-to-one Mathematics tuition is for.