Solutions: Angles of Elevation, Depression and Bearings

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1. Fluency Show Answer

   Horizontal distance A = 40 m, elevation angle θ = 52°. Height O = unknown.

   \[\tan 52° = \frac{O}{40}\] \[O = 40 \times \tan 52° = 40 \times 1.2799\ldots \approx \mathbf{51.20 \text{ m}}\]
2. Fluency Show Answer

   Cliff height O = 75 m, angle of depression = 18°. The angle in the right triangle at the top is 18° (alternate angles). Want horizontal distance A.

   \[\tan 18° = \frac{75}{A}\] \[A = \frac{75}{\tan 18°} = \frac{75}{0.3249\ldots} \approx \mathbf{230.8 \text{ m}}\]
3. Fluency Altitude 3000 m, elevation angle 14°.
   1. Show Answer

      O = 3000 m (height), angle = 14°. Want A (horizontal distance below plane).

      \[\tan 14° = \frac{3000}{A}\] \[A = \frac{3000}{\tan 14°} = \frac{3000}{0.2493\ldots} \approx \mathbf{12\,031 \text{ m} \approx 12.03 \text{ km}}\]
   2. Show Answer

      Slant distance = H. O = 3000 m, angle = 14°.

      \[\sin 14° = \frac{3000}{H}\] \[H = \frac{3000}{\sin 14°} = \frac{3000}{0.2419\ldots} \approx \mathbf{12\,402 \text{ m} \approx 12.40 \text{ km}}\]
4. Understanding Show Answer

   Let h = height of tower, d = initial horizontal distance.

   From initial position: tan 34° = h/d  ⇒  h = d tan 34°  …(1)

   After moving 20 m closer: tan 49° = h/(d − 20)  ⇒  h = (d − 20) tan 49°  …(2)

   Equate (1) and (2):

   \[d\tan 34° = (d - 20)\tan 49°\] \[d \times 0.6745 = d \times 1.1504 - 20 \times 1.1504\] \[d(1.1504 - 0.6745) = 23.008\] \[d \times 0.4759 = 23.008\] \[d = \frac{23.008}{0.4759} \approx 48.35 \text{ m}\]

   Height: h = 48.35 × tan 34° = 48.35 × 0.6745 ≈ 32.6 m

5. Understanding Show Answer

   Lighthouse height = 45 m. Let d₁ = horizontal distance to far boat (angle 24°), d₂ = horizontal distance to near boat (angle 38°).

   \[d_1 = \frac{45}{\tan 24°} = \frac{45}{0.4452} \approx 101.1 \text{ m}\] \[d_2 = \frac{45}{\tan 38°} = \frac{45}{0.7813} \approx 57.6 \text{ m}\]

   Distance between boats = d₁ − d₂ = 101.1 − 57.6 ≈ 43.5 m

6. Understanding
   1. Show Answer

      N 40° E: 40° clockwise from North = 040°

   2. Show Answer

      S 65° W: start at South (180°), rotate 65° toward West = 180° + 65° = 245°

   3. Show Answer

      125° is between 090° (E) and 180° (S). It is 125° − 90° = 35° south of east, so: S 55° E

      (Alternatively: 180° − 125° = 55° from South toward East.)

   4. Show Answer

      305° is between 270° (W) and 360° (N). 360° − 305° = 55° from North toward West, so: N 55° W

7. Understanding Show Answer

   Bearing 072°, distance 85 km.

   \[\text{Northing} = 85\cos 72° = 85 \times 0.3090 \approx \mathbf{26.3 \text{ km north}}\] \[\text{Easting} = 85\sin 72° = 85 \times 0.9511 \approx \mathbf{80.8 \text{ km east}}\]
8. Understanding
   1. Show Answer

      Bearing of B from A = 130°. Bearing of C from A = 220°.

      Angle BAC = 220° − 130° = 90°

      (The two bearings differ by exactly 90°, so the angle at A between AB and AC is 90°.)

   2. Show Answer

      Since angle BAC = 90°, use Pythagoras:

      \[BC = \sqrt{AB^2 + AC^2} = \sqrt{15^2 + 22^2} = \sqrt{225 + 484} = \sqrt{709} \approx \mathbf{26.6 \text{ km}}\]

      Alternatively, cosine rule: BC² = 15² + 22² − 2(15)(22)cos90° = 225 + 484 − 0 = 709. Same answer. ✓

9. Problem Solving Observers A and B, 200 m apart. Elevations: A = 42°, B = 58°.
   1. Show Answer

      Let h = height, d = horizontal distance from A to below aircraft.

      From A: tan 42° = h/d  ⇒  h = d tan 42°  …(1)

      From B: tan 58° = h/(d − 200)  ⇒  h = (d − 200) tan 58°  …(2)

      Equating:

      \[d \times 0.9004 = (d - 200) \times 1.6003\] \[0.9004d = 1.6003d - 320.06\] \[320.06 = 0.6999d\] \[d = \frac{320.06}{0.6999} \approx 457.3 \text{ m from A}\] \[d - 200 \approx 257.3 \text{ m from B}\]

      Horizontal distance from A: 457.3 m; from B: 257.3 m

   2. Show Answer
      \[h = d \times \tan 42° = 457.3 \times 0.9004 \approx \mathbf{411.7 \text{ m}}\]
   3. Show Answer

      Verify from B: h = 257.3 × tan 58° = 257.3 × 1.6003 ≈ 411.7 m ✓

      Both calculations give h ≈ 411.7 m, confirming the result.

10. Problem Solving Hiker: 8 km north, then 11 km on bearing 125°.
    1. Show Answer

       East component of second leg = 11 sin 125° = 11 × sin 55° = 11 × 0.8192 ≈ 9.01 km east

    2. Show Answer

       North component of second leg = 11 cos 125° = 11 × (−cos 55°) = 11 × (−0.5736) ≈ −6.31 km (i.e., 6.31 km south)

    3. Show Answer

       Total northing = 8 + (−6.31) = +1.69 km north

    4. Show Answer

       Total easting = 0 + 9.01 = 9.01 km east

    5. Show Answer

       Direct distance back to start:

       \[D = \sqrt{1.69^2 + 9.01^2} = \sqrt{2.856 + 81.18} = \sqrt{84.04} \approx \mathbf{9.17 \text{ km}}\]

       Bearing back to start: the hiker is 1.69 km N and 9.01 km E of start, so start is 1.69 km S and 9.01 km W of end position.

       \[\text{Bearing} = 180° + \tan^{-1}\!\left(\frac{9.01}{1.69}\right) = 180° + \tan^{-1}(5.33) = 180° + 79.4° \approx \mathbf{259°}\]

       Return distance: 9.17 km on bearing 259° (approximately due west with slight south component).