The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.
Let the height of the tower be h and RQ = x m
Given that $P R=50 \mathrm{~m}$
and $\angle S P Q=30^{\circ}, \angle S R Q=60^{\circ}$
Now, in $\triangle S R Q$; $\tan 60^{\circ}=\frac{S Q}{R Q}$
$\Rightarrow$ $\sqrt{3}=\frac{h}{x} \Rightarrow x=\frac{h}{\sqrt{3}}$ ......(i)
and in $\triangle S P Q, \quad \tan 30^{\circ}=\frac{S Q}{P Q}=\frac{S Q}{P R+A Q}=\frac{h}{50+x}$
$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{50+x}$
$\Rightarrow \quad \sqrt{3} \cdot h=50+x$
$\Rightarrow \quad \sqrt{3} \cdot h=50+\frac{h}{\sqrt{3}}$ [from Eq. (i)]
$\Rightarrow \quad\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) h=50$
$\Rightarrow \quad \frac{(3-1)}{\sqrt{3}} h=50$
$\therefore$ $h=\frac{50 \sqrt{3}}{2}$
$h=25 \sqrt{3} m$
Hence, the required height of tower is 25√3 m.