Finding Unknown Angles Using Trigonometry — Solutions

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1. Find angle θ

   1. O=4, H=8: θ = sin⊃⁻¹(4/8) = sin⊃⁻¹(0.5) = 30.0°
   2. A=6, H=10: θ = cos⊃⁻¹(6/10) = cos⊃⁻¹(0.6) = 53.1°
   3. O=7, A=7: θ = tan⊃⁻¹(7/7) = tan⊃⁻¹(1) = 45.0°
   4. O=3, H=5: θ = sin⊃⁻¹(3/5) = sin⊃⁻¹(0.6) = 36.9°
   5. A=9, H=15: θ = cos⊃⁻¹(9/15) = cos⊃⁻¹(0.6) = 53.1°
   6. O=12, A=5: θ = tan⊃⁻¹(12/5) = tan⊃⁻¹(2.4) = 67.4°
   7. O=8, H=17: θ = sin⊃⁻¹(8/17) = sin⊃⁻¹(0.4706) = 28.1°
   8. A=4.5, H=9: θ = cos⊃⁻¹(4.5/9) = cos⊃⁻¹(0.5) = 60.0°
   9. O=11, A=14: θ = tan⊃⁻¹(11/14) = tan⊃⁻¹(0.7857) = 38.1°
   10. O=6.2, H=13.0: θ = sin⊃⁻¹(6.2/13.0) = sin⊃⁻¹(0.4769) = 28.5°
   11. A=20, H=25: θ = cos⊃⁻¹(20/25) = cos⊃⁻¹(0.8) = 36.9°
   12. O=15, A=8: θ = tan⊃⁻¹(15/8) = tan⊃⁻¹(1.875) = 61.9°

2. Identify which inverse to use and verify

   1. Liam’s approach:
      1. Liam is wrong. The sides given are opposite (8) and adjacent (15). These are connected by tangent, not cosine. The correct inverse is tan⊃⁻¹(8/15).
      2. θ = tan⊃⁻¹(8/15) = tan⊃⁻¹(0.5333) = 28.1°
   2. O=9, H=41:
      1. θ = sin⊃⁻¹(9/41) = sin⊃⁻¹(0.2195) = 12.7°
      2. Adjacent = 41 × cos(12.7°) = 41 × 0.9756 = 40.0
      3. Check: 9² + 40² = 81 + 1600 = 1681 = 41² ✓
   3. Inverse trig in 45°-45°-90° triangle: In a 45°-45°-90° triangle with legs = 1 and hypotenuse = √2: sin⊃⁻¹(1/√2) = 45° and cos⊃⁻¹(1/√2) = 45°. This occurs because the two legs are equal, making both acute angles identical at 45°. Both inverse functions give the same answer here.
   4. 3-4-5 triangle scaled to 5-12-13; angle opposite side 5: sin θ = 5/13, cos θ = 12/13, tan θ = 5/12. Using sin⊃⁻¹: θ = sin⊃⁻¹(5/13) = sin⊃⁻¹(0.3846) = 22.6°

3. Elevation and depression problems

   1. Building angle of elevation:
      1. Right triangle: horizontal base (adj = 30 m), vertical building height (opp = 45 m), angle of elevation θ at the observer’s feet.
      2. θ = tan⊃⁻¹(45/30) = tan⊃⁻¹(1.5) = 56.3°
   2. Lifeguard angle of depression: tan α = 8/22 → α = tan⊃⁻¹(8/22) = tan⊃⁻¹(0.3636) = 20.0°
   3. Ski slope angle: θ = tan⊃⁻¹(420/1260) = tan⊃⁻¹(0.3333) = 18.4°
   4. Cliff and two boats (cliff = 60 m):
      1. Boat A: tan(25°) = 60/d⊂A; d⊂A; = 60 ÷ tan(25°) = 60 ÷ 0.4663 = 128.7 m
      2. Boat B: tan(40°) = 60/d⊂B; d⊂B; = 60 ÷ tan(40°) = 60 ÷ 0.8391 = 71.5 m
      3. Distance between boats = 128.7 − 71.5 = 57.2 m

4. Car driving up a hill (rise 8 m, run 60 m; then rise 5 m, run 35 m)

   1. Right triangle: horizontal = 60 m (adjacent), vertical = 8 m (opposite), road = hypotenuse.
   2. θ = tan⊃⁻¹(8/60) = tan⊃⁻¹(0.1333) = 7.6°
   3. Second section: θ = tan⊃⁻¹(5/35) = tan⊃⁻¹(0.1429) = 8.1°. The second section is steeper (8.1° > 7.6°).

5. Roof rafter pitch angle (height 3.2 m, run 6 m)

   1. θ = tan⊃⁻¹(3.2/6) = tan⊃⁻¹(0.5333) = 28.1°
   2. Complementary angle = 90° − 28.1° = 61.9°. Error introduced = 61.9° − 28.1° = 33.8°. This is a significant error that would make the rafter cut completely wrong.

6. Tower angles of elevation from 20 m and 50 m (tower height = 28 m)

   1. From Point A: θ = tan⊃⁻¹(28/20) = tan⊃⁻¹(1.4) = 54.5°
   2. From Point B: θ = tan⊃⁻¹(28/50) = tan⊃⁻¹(0.56) = 29.2°
   3. Difference = 54.5° − 29.2° = 25.3°

7. Plane flight bearing (240 km east, 180 km north)

   1. Right triangle with eastward leg = 240 km and northward leg = 180 km. The angle at the origin is measured from north (vertical) toward east.
   2. The angle from north: tan(θ) = east/north = 240/180 = 1.3333 → θ = tan⊃⁻¹(1.3333) = 53.1°
   3. Bearing = N53.1°E, or equivalently 053.1° (measured clockwise from north).

8. Triangular plot (right angle at C, CA = 45 m, CB = 60 m)

   1. Angle A: the side opposite A is CB = 60 m, adjacent is CA = 45 m. tan(A) = 60/45 = 1.3333 → A = tan⊃⁻¹(1.3333) = 53.1°
   2. Angle B = 90° − 53.1° = 36.9°. Verify: tan(B) = 45/60 = 0.75 → tan⊃⁻¹(0.75) = 36.9° ✓
   3. AB = √(45² + 60²) = √(2025 + 3600) = √5625 = 75.00 m

9. Coastguard on cliff (35 m high, yacht 120 m away)

   1. α = tan⊃⁻¹(35/120) = tan⊃⁻¹(0.2917) = 16.2° (angle of depression)
   2. The rescue boat travels horizontally 120 m. The angle it makes with the base of the cliff is the same as the angle of depression (alternate interior angles with the horizontal). θ = tan⊃⁻¹(35/120) = 16.2°
   3. Yes — both answers are 16.2°. This confirms the alternate interior angles theorem: the angle of depression equals the angle of elevation from the same two points.

10. Architect’s triangular wall (base 8 m, height 5 m)

    1. Angle at bottom-right: tan(θ) = opposite/adjacent = 5/8 = 0.625 → θ = tan⊃⁻¹(0.625) = 32.0°
    2. The support beam runs from the bottom-right corner (8 m along base from left) to a point on the left vertical side 2 m above the floor. Horizontal distance = 8 m, vertical distance = 2 m. Angle = tan⊃⁻¹(2/8) = tan⊃⁻¹(0.25) = 14.0°
    3. Length = √(8² + 2²) = √(64 + 4) = √68 = 8.25 m. Or: using trig: beam = 2 ÷ sin(14.0°) = 2 ÷ 0.2419 = 8.27 m (slight difference due to rounding). Pythagoras is exact: 8.25 m.