These eight questions use new numbers and ramp from choosing a rule to a two-triangle case and a 3D problem. Write your own working first, then open the answer.
They follow the lessons in solution of triangles. Give lengths and areas to three significant figures and angles to one decimal place.
Questions
Question 1. Which rule would you use to start each triangle?
- (a) a = 7, b = 9, C = 50°
- (b) A = 40°, B = 65°, a = 8
- (c) a = 5, b = 6, c = 7
- (d) A = 30°, a = 5, b = 8
Answer
- (a) SAS, so the cosine rule.
- (b) Two angles and a side, so the sine rule.
- (c) SSS, so the cosine rule.
- (d) Two sides and an angle opposite one of them, so the sine rule, and check the ambiguous case.
Question 2. In triangle ABC, AB = 6 cm, AC = 10 cm and angle A = 110°. Find BC.
Answer
The angle is between the two sides, so use the cosine rule. BC² = 6² + 10² − 2(6)(10) cos 110° = 136 − 120(−0.3420) = 136 + 41.04 = 177.04.
BC = √177.04 ≈ 13.3 cm. A negative cosine for an obtuse angle makes the side longer, which is correct.
Question 3. In triangle ABC, angle A = 52°, angle B = 71° and a = 15 cm. Find b.
Answer
Side a is opposite A, so the pair (52°, 15 cm) is known. b ÷ sin 71° = 15 ÷ sin 52°.
b = 15 × 0.9455 ÷ 0.7880 ≈ 18.0 cm.
Question 4. A triangle has sides 8 cm, 11 cm and 14 cm. Find its largest angle.
Answer
The largest angle is opposite the longest side, 14 cm. cos C = (8² + 11² − 14²) ÷ (2 × 8 × 11) = (64 + 121 − 196) ÷ 176 = −11 ÷ 176 = −0.0625.
C ≈ 93.6°. The negative cosine confirms that the angle is obtuse.
Question 5. In triangle ABC, angle A = 25°, a = 6 cm and b = 10 cm. Find the possible values of angle B.
Answer
sin B = 10 sin 25° ÷ 6 = 4.226 ÷ 6 ≈ 0.7044, so B ≈ 44.8° or 180° − 44.8° = 135.2°.
Test: 25° + 135.2° = 160.2° < 180°, so both fit. B = 44.8° or 135.2°.
Question 6. In triangle ABC, angle A = 40°, a = 4 cm and b = 9 cm. How many triangles can be drawn?
Answer
sin B = 9 sin 40° ÷ 4 = 5.785 ÷ 4 ≈ 1.446, which is greater than 1. No angle has that sine, so no triangle exists. Equivalently, a = 4 is less than b sin A = 5.79.
Question 7. In triangle ABC, AB = 15 cm, angle A = 48° and angle B = 62°. Find the area.
Answer
Angle C = 180° − 48° − 62° = 70°. Sine rule: BC ÷ sin 48° = 15 ÷ sin 70°, so BC = 15 × 0.7431 ÷ 0.9397 ≈ 11.863 cm.
AB and BC enclose angle B = 62°, so area = ½ × 15 × 11.863 × sin 62° = 88.97 × 0.8829 ≈ 78.6 cm².
Question 8. Points A and B are on level ground, 80 m apart. A vertical tower PT has its foot at P.
Angle PAB = 35° and angle PBA = 80°, and the angle of elevation of T from A is 20°. Find the height of the tower.
Answer
Ground triangle ABP: angle APB = 180° − 35° − 80° = 65°. AP ÷ sin 80° = 80 ÷ sin 65°, so AP = 80 × 0.9848 ÷ 0.9063 ≈ 86.93 m.
Vertical triangle APT: PT = AP tan 20° = 86.93 × 0.3640 ≈ 31.6 m.
If you got these wrong
Wrong start in Question 1 or 2: revisit choosing sine rule or cosine rule. Missing answers in Questions 5 and 6: read the ambiguous sine-rule case.
Area slips in Question 7 point to triangle area without the included angle. Trouble with Question 8 points to solving 3D triangle problems. Log each slip in the mistake log, and try a timed practice session when you are ready.
If you want a teacher to go through your working, see online one-to-one Additional Mathematics tuition.