Question

A parachutist is descending vertically and makes angles of elevation of 45 degree and 60 degree at two observing points 100 m apart from each other on the left side of himself .find the maximum height from which he falls and the distance of the point where he falls on the ground from the just observation point

Answer 1

hey I am solve on paper hope this is help you

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

$tan \theta = \frac{Perpendicular}{Base}$

$tan 60^{\circ}= \frac{AB}{BC}$

$\sqrt{3}= \frac{h}{x}$

$\sqrt{3}x= h$  ---1

In ΔABD

$tan \theta = \frac{Perpendicular}{Base}$

$tan 45^{\circ}= \frac{AB}{BD}$

$1= \frac{h}{100+x}$

$100+x= h$  ---2

Equating 1 and 2

$\sqrt{3}x= 100+x$

$\sqrt{3}x-x= 100$

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

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

$x= 136.602$

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