Question

The shadow of a tower is found to be $60$ ft. longer when the suns altitude has become $60$ from $30$. The height of the tower from the ground is

• A. $350$ ft. app.
• B. $400$ ft. app.
• C. $52$ft. app.
• D. None

Answer 1

The correct option is C $52$ft. app.

As shown in the figure let $A$ be the position of the sun.

Let height of tower= $h$

Shadow of tower falls along the segment $BD$

Let $BC=xft$ and $DC=60ft$

In $△ABC,\tan {60}^o=\frac{h}{x}$

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

In $△ABD,\tan {30}^o=\frac{h}{60+x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{60+x}$

$=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}x}{60+x}\Rightarrow 60+x=3x$

$\Rightarrow x=30ft$

$\Rightarrow h=\sqrt{3}\times 30ft$

$=51.96ft\simeq 52ft$ .