A parachutist is descending vertically and makes angles of elevation of 45° and 60°

Solution:

Let BC be the height h of the parachutist and makes an angle of elevations and  respectively at two observing points  apart from each other.

So we use trigonometric ratios.

In triangle BCD

$\tan 60^{\circ}=\frac{h}{x}$

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

Now in triangle,

$\tan 45^{\prime}=\frac{h}{x+100}$

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

$\Rightarrow x+100=h$

$\Rightarrow \frac{h}{\sqrt{3}}+100=h$

$\Rightarrow h+100 \sqrt{3}=\sqrt{3} h$

$\Rightarrow h=\frac{100 \sqrt{3}}{\sqrt{3}-1}$

$\Rightarrow h=50(3+\sqrt{3})$

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

$x=\frac{50(3+\sqrt{3})}{\sqrt{3}}$

$=50(1+\sqrt{3})$

Hence the maximum height is $50(3+\sqrt{3}) \mathrm{m}=236.6 \mathrm{~m}$. and distance is $50(1+\sqrt{3}) \mathrm{m}=136.6 \mathrm{~m}$.