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 )​

Answer 1

Answer:

Let AC represent the initial height and point A represent the initial position of the plane.  Let point B represent the position where plane lands.  Angle of depression = ∠EAB = 20° Now, seg AE || seg BC ∴ ∠ABC = ∠EAB … [Alternate angles]  ∴ ∠ABC = 20°  Speed of the plane = 200 km/hr  = 200 × 1000/3600 m/sec  = 500/9 m/sec  ∴ Distance travelled in 54 sec = speed × time  = 500/9 × 54 = 3000 m  ∴ AB = 3000 m  In right angled ∆ABC, sin 20° = AC/AB  ….[By definition]  ∴ 0.342 = AC/3000 ∴ AC = 0.342 × 3000 = 1026 m  ∴ The plane was at a height of 1026 m when it started landing.Read more on Sarthaks.com - https://www.sarthaks.com/858044/while-landing-airport-pilot-made-angle-depression-20-average-speed-the-plane-was-200-plane?show=858049#a858049

Answer 2

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

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