Question

Question 7
The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is ${30}^{\circ }$ than when it is ${60}^{\circ }$. Find the height of the tower.

Answer 1

Let the height of the tower be h and RQ = x m.

Given that, PR = 50 m

And, $\angle SPQ={30}^{\circ },\angle SRQ={60}^{\circ }$

Now, in $\Delta SRQ$, $tan{60}^{\circ }=\frac{SQ}{RQ}$

$\Rightarrow \sqrt{3}=\frac{h}{x}\Rightarrow x=\frac{h}{\sqrt{3}}$ ...(i)

And, in $\Delta SPQ,tan{30}^{\circ }=\frac{SQ}{PQ}=\frac{SQ}{PR+RQ}=\frac{h}{50+x}$



$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{50+x}$

$\Rightarrow \sqrt{3}.h=50+x$

$\Rightarrow \sqrt{3}.h=50+\frac{h}{\sqrt{3}}$ [from Eq. (i)]

$\Rightarrow (\sqrt{3}-\frac{1}{\sqrt{3}})h=50$

$\Rightarrow \frac{(3-1)}{\sqrt{3}}h=50$

$\therefore h=\frac{50\sqrt{3}}{2}h$

Hence, the required height of the tower is $25\sqrt{3}m$.