Question

A parachutist is descending vertically and makes angles of elevation of 45° and 60° at two observing points 100 metres apart from each other on the left side of him. Find the maximum height from which he falls and the distance of
the point where he falls on the ground from the second point of observation.
[HOTS] ​

Answer 1

The maximum height from which he falls is 236.602 m and the distance of the point where he falls on the ground from the just observation point is 136.602 m

• Let the height of the parachutist be h

∠ACB = 60°

∠ADC =45°

DC = 100 m

• Hence

The maximum height from which he falls is 236.602 m and the distance of the point where he falls on the ground from the just observation point is 136.602 m

Answer 2

Answer:

The maximum height from which he falls is 236.602 m and the distance of the point where he falls on the ground from the just observation point is 136.602 m

Step-by-step explanation:

Let the height of the parachutist be h

∠ACB = 60°

∠ADC =45°

DC = 100 m

In ΔABC

$\tan( \alpha ) = \frac{perpendiular}{base} \\ \\ \tan( 60 ) = \frac{ab}{bc} \\ \\ \sqrt{3} = \frac{h}{x} \\ \\ h \: = x \sqrt{3}$

 ---1

In ΔABD

$\tan( \alpha ) = \frac{perpendiular}{base} \\ \\ \tan( 45 ) = \frac{ab}{bd} \\ \\ 1 = \frac{h}{100 + x} \\ \\ h \: = 100 + x$

 ---2

Equating 1 and 2

$\sqrt{3} = 100 + x \\ \\ x\sqrt{3} - x = 100 \\ \\ x( \sqrt{3} - 1) = 100 \\ \\ x = \frac{100}{ \sqrt{3} - 1 } \\ \\ x = 136.620$

So, Height of parachutist = h = 100+x=100+136.602=236.602 m

Hence The maximum height from which he falls is 236.602 m and the distance of the point where he falls on the ground from the just observation point is 136.602 m