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

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.