Topic Review — Trigonometry — Solutions

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This review covers all lessons in Trigonometry: the sine rule (including the ambiguous case), the cosine rule and area formula, and 2D and 3D trigonometry applications. Questions are mixed in difficulty.

Review Questions

1. In triangle ABC, A = 42°, B = 67°, and a = 11 cm. Find the length of side b. ▶ Show Answer
   C = 180 − 42 − 67 = 71°
   b/sin 67° = 11/sin 42°
   b = 11 × sin 67°/sin 42° = 11 × 0.9205/0.6691 ≈ 15.1 cm
2. In triangle ABC, a = 8, b = 12, and A = 38°. Find all possible values of angle B, checking both solutions for validity. ▶ Show Answer
   sin B = b × sin A/a = 12 × sin 38°/8 = 12 × 0.6157/8 = 0.9236
   B₁ = sin⁻¹(0.9236) ≈ 67.4°; C₁ = 180 − 38 − 67.4 = 74.6° > 0° ✓
   B₂ = 180 − 67.4 = 112.6°; C₂ = 180 − 38 − 112.6 = 29.4° > 0° ✓
   Ambiguous case: B ≈ 67.4° or B ≈ 112.6°
3. In triangle XYZ, X = 105°, x = 20 m, and y = 9 m. Explain why there is only one possible triangle and find angle Y. ▶ Show Answer
   Since X = 105° is obtuse, no other angle can be obtuse (angle sum would exceed 180°), so Y must be acute — only one solution.
   sin Y = y × sin X/x = 9 × sin 105°/20 = 9 × 0.9659/20 = 0.4347
   Y = sin⁻¹(0.4347) ≈ 25.8°
4. Find the area of triangle PQR where PQ = 14 m, PR = 10 m, and angle QPR = 72°. ▶ Show Answer
   Area = ½ × PQ × PR × sin(QPR)
   = ½ × 14 × 10 × sin 72°
   = 70 × 0.9511
   ≈ 66.6 m²
5. In triangle ABC, the area is 30 cm², side a = 9 cm, and side b = 8 cm. Find all possible values of angle C, and for each valid case find side c. ▶ Show Answer
   Area = ½ab sin C: 30 = ½ × 9 × 8 × sin C = 36 sin C
   sin C = 30/36 = 5/6 ≈ 0.8333
   C₁ ≈ 56.4°: c² = 81 + 64 − 144 cos 56.4° = 145 − 144 × 0.5519 = 145 − 79.5 = 65.5; c ≈ 8.1 cm
   C₂ ≈ 123.6°: c² = 81 + 64 − 144 cos 123.6° = 145 − 144 × (−0.5519) = 145 + 79.5 = 224.5; c ≈ 15.0 cm
   Both are valid: C ≈ 56.4° (c ≈ 8.1 cm) or C ≈ 123.6° (c ≈ 15.0 cm)
6. In triangle ABC, a = 7 cm, b = 9 cm, and C = 65°. Find side c using the cosine rule. ▶ Show Answer
   c² = a² + b² − 2ab cos C
   = 49 + 81 − 2 × 7 × 9 × cos 65°
   = 130 − 126 × 0.4226
   = 130 − 53.25 = 76.75
   c = √76.75 ≈ 8.76 cm
7. In triangle ABC, a = 10, b = 7, c = 6. Find the largest angle. ▶ Show Answer
   The largest angle is opposite the longest side (a = 10).
   cos A = (b² + c² − a²)/(2bc) = (49 + 36 − 100)/(2 × 7 × 6) = −15/84 = −0.1786
   A = cos⁻¹(−0.1786) ≈ 100.3°
8. Two ships leave port at the same time. Ship A travels 24 km on a bearing of 050°. Ship B travels 18 km on a bearing of 140°. Find the distance between the two ships. ▶ Show Answer
   Angle between the two bearings at port = 140 − 50 = 90°.
   Since the included angle is 90°, use Pythagoras:
   d = √(24² + 18²) = √(576 + 324) = √900 = 30 km
9. Three towns A, B, C form a triangle. AB = 60 km, BC = 45 km, AC = 50 km. Find the angle at B (angle ABC). ▶ Show Answer
   Using cosine rule to find angle B (sides from B are BA = 60 and BC = 45, side opposite B is AC = 50):
   cos B = (BA² + BC² − AC²)/(2 × BA × BC)
   = (3600 + 2025 − 2500)/(2 × 60 × 45)
   = 3125/5400 = 0.5787
   B = cos⁻¹(0.5787) ≈ 54.6°
10. A triangular garden has sides of length 12 m, 15 m, and 20 m. Find the area of the garden using Heron’s formula or by first finding an angle. ▶ Show Answer
    Find angle C opposite the longest side (c = 20):
    cos C = (12² + 15² − 20²)/(2 × 12 × 15) = (144 + 225 − 400)/360 = −31/360 = −0.0861
    C = cos⁻¹(−0.0861) ≈ 94.94°
    Area = ½ × 12 × 15 × sin 94.94° = 90 × sin 94.94° = 90 × 0.9963 ≈ 89.7 m²
11. From the top of a 60 m cliff, the angle of depression to a boat is 28°. Find the horizontal distance from the cliff base to the boat, and the direct distance from the top of the cliff to the boat. ▶ Show Answer
    Horizontal distance: tan 28° = 60/d ⇒ d = 60/tan 28° = 60/0.5317 ≈ 112.9 m
    Direct distance (hypotenuse): d_slant = 60/sin 28° = 60/0.4695 ≈ 127.8 m
12. A rectangular box has length 10 m, width 8 m, and height 5 m. Find the angle the space diagonal makes with the base of the box. ▶ Show Answer
    Base diagonal = √(10² + 8²) = √164 ≈ 12.806 m
    The space diagonal forms a right triangle with the base diagonal (adjacent) and height (opposite).
    tan θ = 5/12.806 = 0.3904
    θ = tan⁻¹(0.3904) ≈ 21.4°
13. A yacht sails 20 km on a bearing of 025° from port P to point A, then 15 km on a bearing of 115° from A to point B. Find the distance PB and the bearing of B from P. ▶ Show Answer
    Back-bearing of PA from A = 025 + 180 = 205°; bearing AB = 115°.
    Interior angle at A = 205 − 115 = 90° (right angle at A)
    PB = √(20² + 15²) = √(400 + 225) = √625 = 25 km
    sin(∠APB) = 15/25 = 0.6 ⇒ ∠APB = 36.87°
    Bearing of B from P = 025 + 36.9 ≈ 062°
14. A right pyramid has a rectangular base 8 m × 6 m and a vertical height of 10 m. Find the slant height from the apex to the midpoint of the longer base edge, and the angle this slant makes with the base. ▶ Show Answer
    The midpoint of the longer edge (8 m) is 3 m from the centre (half of 6 m width).
    Slant height = √(3² + 10²) = √(9 + 100) = √109 ≈ 10.44 m
    Angle with base: tan θ = 10/3 = 3.333
    θ = tan⁻¹(3.333) ≈ 73.3°
15. Three surveying stations A, B, and C lie in a horizontal plane. From A, the bearing to B is 070° and the distance AB = 200 m. From A, the bearing to C is 140° and AC = 160 m. A vertical mast of height 45 m stands at B. Find the area of triangle ABC and the angle of elevation of the top of the mast from C. ▶ Show Answer
    Angle BAC: bearings from A are 070° and 140°, so angle BAC = 140 − 70 = 70°
    
    Area of triangle ABC:
    Area = ½ × AB × AC × sin(BAC) = ½ × 200 × 160 × sin 70°
    = 16000 × 0.9397 ≈ 15 035 m²
    
    Distance BC (horizontal) using cosine rule:
    BC² = 200² + 160² − 2 × 200 × 160 × cos 70°
    = 40000 + 25600 − 64000 × 0.3420
    = 65600 − 21888 = 43712
    BC = √43712 ≈ 209.1 m
    
    Angle of elevation of mast top from C:
    tan θ = 45/209.1 = 0.2152
    θ = tan⁻¹(0.2152) ≈ 12.1°