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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.
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