Applications of Trigonometry — Solutions

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1. Elevation and depression

   1. Building height; elev = 34°, dist = 50 m: tan(34°) = h/50 → h = 50×tan(34°) = 50×0.6745 = 33.73 m
   2. Kite string; ht = 40 m, elev = 52°: sin(52°) = 40/L → L = 40÷sin(52°) = 40÷0.7880 = 50.76 m
   3. Boat from cliff; cliff = 20 m, dep = 15°: tan(15°) = 20/d → d = 20÷tan(15°) = 20÷0.2679 = 74.64 m
   4. Tower; dist = 80 m, elev = 27°: tan(27°) = h/80 → h = 80×0.5095 = 40.76 m
   5. Plane; alt = 1500 m, dep = 3°: tan(3°) = 1500/d → d = 1500÷tan(3°) = 1500÷0.05241 = 28 625.45 m ≈ 28.6 km
   6. Car from window; ht = 12 m, dep = 40°: tan(40°) = 12/d → d = 12÷0.8391 = 14.30 m

2. Bearings and direction problems

   1. Ship bearing 040°, 120 km:
      1. Right triangle with hypotenuse = 120 km, angle 40° from the north (vertical). North component is adjacent, East component is opposite.
      2. Northward = 120×cos(40°) = 120×0.7660 = 91.92 km
      3. Eastward = 120×sin(40°) = 120×0.6428 = 77.13 km
   2. Helicopter bearing 310°, 85 km: 310° = N50°W, so angle from north = 50°. Northward = 85×cos(50°) = 85×0.6428 = 54.6 km. Westward = 85×sin(50°) = 85×0.7660 = 65.1 km.
   3. Hiker A→B→C:
      1. AC = √(6.4² + 4.8²) = √(40.96 + 23.04) = √64 = 8 km
      2. The bearing from A to C: the angle east of north = tan⊃⁻¹(6.4/4.8) = tan⊃⁻¹(1.3333) = 53.1°. Bearing = 053.1°
   4. Two ships, angle between bearings 060° and 150° = 90°: The angle between the two bearings is 150° − 60° = 90°. So the two ships form a right-angle triangle at the port. Distance = √(50² + 50²) = 50√2 = 70.71 km

3. Multi-step and mixed applications

   1. Two buildings:
      1. Building A: tan(58°) = h⊂A;/15 → h⊂A; = 15×tan(58°) = 15×1.6003 = 24.00 m
      2. Building B: tan(32°) = h⊂B;/22 → h⊂B; = 22×tan(32°) = 22×0.6249 = 13.75 m
      3. Difference = 24.00 − 13.75 = 10.25 m
   2. Hill problem:
      1. Vertical height = 300×sin(22°) = 300×0.3746 = 112.38 m
      2. Horizontal distance = 300×cos(22°) = 300×0.9272 = 278.16 m
      3. Descent path: sin(18°) = 112.38/L → L = 112.38÷sin(18°) = 112.38÷0.3090 = 363.69 m
   3. Satellite dish:
      1. Distance = √(9² + 12²) = √(81+144) = √225 = 15 m
      2. Angle = tan⊃⁻¹(9/12) = tan⊃⁻¹(0.75) = 36.9°
      3. Height above dish over 2 m horizontal: tan(65°) = h/2 → h = 2×tan(65°) = 2×2.1445 = 4.29 m
   4. Building roof and car park:
      1. Near edge: tan(55°) = 30/d⊂1; → d⊂1; = 30÷tan(55°) = 30÷1.4281 = 21.01 m
      2. Far edge: tan(28°) = 30/d⊂2; → d⊂2; = 30÷tan(28°) = 30÷0.5317 = 56.42 m
      3. Length of car park = 56.42 − 21.01 = 35.41 m

4. Real-world trigonometry applications

   1. Ladder 5 m, angle 65°: sin(65°) = h/5 → h = 5×sin(65°) = 5×0.9063 = 4.53 m
   2. Ramp 10 m, angle 8°: sin(8°) = h/10 → h = 10×sin(8°) = 10×0.1392 = 1.39 m
   3. Zip-line pole 14 m, horizontal 20 m: tan(θ) = 14/20 = 0.7 → θ = tan⊃⁻¹(0.7) = 35.0° (angle of depression)

5. Bearings and navigation problems

   1. Vessel at bearing 125°, 40 km: 125° = S55°E. South = 40×cos(55°) = 40×0.5736 = 22.9 km. East = 40×sin(55°) = 40×0.8192 = 32.8 km.
   2. Yacht 30 km on bearing 200°:
      1. 200° = S20°W. Southward = 30×cos(20°) = 30×0.9397 = 28.2 km
      2. Westward = 30×sin(20°) = 30×0.3420 = 10.3 km
      3. Return trip is same distance = 30 km on bearing 020° (directly opposite to 200°).
   3. Aircraft A at 070° 200 km, Aircraft B at 340° 150 km: A: east=200×sin(70°)=187.9, north=200×cos(70°)=68.4. B: east=150×sin(340°)=−51.3, north=150×cos(340°)=140.9. Diff: east=187.9−(−51.3)=239.2, north=68.4−140.9=−72.5. Distance=√(239.2²+72.5²)=√(57217+5256)=√62473=250 km

6. Angles of elevation with changing position

   1. Two elevations to lighthouse: tan(20°)=h/d → d=h/tan(20°)=h/0.3640. tan(35°)=h/(d−50) → h/0.7002 = h/0.3640 − 50. Solving: h(1/0.7002 − 1/0.3640)=50; h(1.4282−2.7473)=50; h(−1.3191)=50; h=37.9 m
   2. Cliff 80 m, boat moving: First position: d&sub1; = 80/tan(12°) = 80/0.2126 = 376.3 m. Second position: d&sub2; = 80/tan(25°) = 80/0.4663 = 171.6 m. Distance travelled = 376.3 − 171.6 = 205 m

7. Three-dimensional trigonometry applications

   1. Pool 25 m × 10 m: Diagonal = √(25²+10²) = √(625+100) = √725 = 26.93 m. Angle with longer side = tan⊃⁻¹(10/25) = tan⊃⁻¹(0.4) = 21.80°
   2. Flagpole on hill:
      1. Height of base = 60×tan(15°) = 60×0.2679 = 16.08 m
      2. Total height to top of flagpole = 16.08 + 8 = 24.08 m. Angle = tan⊃⁻¹(24.08/60) = tan⊃⁻¹(0.4013) = 21.87°

8. Multi-step elevation and depression problems

   1. Two towers:
      1. tan(42°) = 30/d → d = 30/tan(42°) = 30/0.9004 = 33.32 m
      2. Additional height = 50×tan(18°) = 50×0.3249 = 16.25 m
      3. Total height of Tower B = 30 + 16.25 = 46.25 m
   2. Surveyor and hill: Let h = height, d = far distance. tan(14°)=h/d and tan(23°)=h/(d−200). From first: d=h/tan(14°)=h/0.2493. Substitute: h/0.4245 = h/0.2493 − 200. h(1/0.4245−1/0.2493)=200; h(2.3557−4.0112)=200; h(−1.6555)=200; h=121 m

9. Return-journey and triangular bearing problems

   1. Helicopter from base:
      1. North = 80×cos(55°)=80×0.5736=45.9 km. East = 80×sin(55°)=80×0.8192=65.5 km.
      2. D = H minus 60 km east: North = 45.9, East = 65.5 − 60 = 5.5. So D is at (5.5, 45.9) relative to base.
      3. Distance = √(5.5²+45.9²) = √(30.25+2106.81) = √2137.06 = 46.2 km. Bearing from D to base: angle = tan⊃⁻¹(5.5/45.9)≈6.8° west of south = bearing 173.2° (S6.8°W).
   2. Triangular race course: B: north=12×cos(30°)=10.4 km, east=12×sin(30°)=6.0 km. C from B: bearing 120°(S60°E), north=−9×sin(30°)=−4.5, east=9×cos(30°)=7.8. So C: north=10.4−4.5=5.9, east=6.0+7.8=13.8. Distance CA=√(5.9²+13.8²)=√(34.81+190.44)=√225.25=15.0 km. Bearing from C to A: angle from south=tan⊃⁻¹(13.8/5.9)=66.8° west of south; bearing=247° (S67°W).

10. Shadow and angle-of-elevation investigations

    1. Sun at 38°, shadow 45 m: tan(38°) = h/45 → h = 45×tan(38°) = 45×0.7813 = 35.16 m
    2. Two posts 4 m and 7 m, 10 m apart: Using similar triangles: crossing height h = (4×7)/(4+7) = 28/11 = 2.55 m above the ground.
    3. Shadow at 25° and 60°, pole 10 m: Shadow at 25°: d&sub1;=10/tan(25°)=10/0.4663=21.45 m. Shadow at 60°: d&sub2;=10/tan(60°)=10/1.7321=5.77 m. Difference = 21.45−5.77 = 15.68 m