#### Question

While landing at an airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing. (sin 20° = 0.342)

#### Solution

Let the plane was at a height of h m when it started landing.

Average speed of the plane = 200 km/h = \[200 \times \frac{5}{18} = \frac{500}{9}\] m/s

Time taken by plane to reach the ground = 54 s

∴ Distance covered by the plane to reach the ground = Average speed of the plane × Time taken by plane to reach the ground

= \[\frac{500}{9} \times 54\]

= 3000 m

Here, AC = 3000 m and ∠ACB = 20º

In right ∆ABC,

\[\sin20^\circ = \frac{AB}{AC}\]

\[ \Rightarrow 0 . 342 = \frac{h}{3000}\]

\[ \Rightarrow h = 0 . 342 \times 3000 = 1026 m\]

Thus, the plane was at the height of 1026 m when it started landing.