The shadow of a tower standing on a level

Question:

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.

Solution:

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.

Leave a comment

Click here to get exam-ready with eSaral