Question

The shadow of a tower at a time is three times as long as its shadow when the angle of elevation is ${60}^{\circ }$. Find the angle of elevation of the sun at a time of the longer shadow.

Answer 1

Let the height of the flagstaff be h m and the angle between the sun rays and the ground at the time of longer shadow be $\theta$.
BC and BD are the lengths of the shadow of the flagstaff when the angle between the sunrays and the ground are ${60}^{\circ }$ and $\theta$ respectively.
Given, BD = 3 BC ...(1)
In $\Delta$ ABC,
$tan{60}^{\circ }=\frac{AB}{BC}$
$\Rightarrow h=\sqrt{3}BC$ ...(2)
In $\Delta$ ABC,
$tan\theta =\frac{AB}{BD}$
$\therefore tan\theta =\frac{h}{BD}$ ...(3)
From (1), (2) and (3), we get
$\sqrt{3}BC=tan\theta \times 3BC$
$\therefore tan\theta =\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$
$\Rightarrow tan\theta =\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$
$\Rightarrow \theta ={30}^{\circ }(\because tan{30}^{\circ }=\frac{1}{\sqrt{3}})$
Thus, the angle between the sunrays and the ground at the time of longer shadow is ${30}^{\circ }$.