Question

The angle of elevation of an aeroplane from A on the grounds is 45°. After 15 seconds flight, the angle of elevation changes to 30°. If the aeroplane is flying at a height of 3000 m, find the speed of the plane.

Answer 1

Given: Angle of elevation at A initially = 45°, after 15s = 30°, height of plane = 3000 m.

Step-wise calculation:

1. Let the horizontal distance from A to the plane when angle = 45° be d₁ and when angle = 30° be d₂.

2. Using `tan θ = "Opposite"/"Adjacent"`:

`tan 45^circ = 1 = "Height"/d_1` 

⇒ d₁ = height = 3000 m

`tan 30^circ = 1/sqrt(3)`

= `"Height"/d_2`

⇒ `d_2 = "Height" xx sqrt(3)`

= `3000sqrt(3)` m

3. Horizontal distance travelled in 15s

= d₂ – d₁

= `3000(sqrt(3) - 1)` m

4. Speed = `"Distance"/"Time"`

= `(3000(sqrt(3) - 1))/15`

= `200(sqrt(3) - 1)` m/s

5. Numeric value: `sqrt(3) ≈ 1.7320508`

⇒ Speed = 200 × 0.7320508

= 146.41 m/s 

If wanted in km/h: 146.41 × 3.6 = 527.08 km/h.

Speed of the plane = `200(sqrt(3) - 1)` m/s ≈ 146.41 m/s (≈ 527.08 km/h).