Question

A vertical tower of height $50$ meters high stands on a sloping ground. The bottom of the tower is at the same level as the middle point of a vertical flagpole. From the top of the tower, the angles of depression of the top and bottom of the flagpole are ${15}^0$ and ${45}^0$ respectively. The height of the flagpole is

• A. $\frac{50}{\sqrt{3}}m$
• B. $50\sqrt{3}m$
• C. $\frac{100}{\sqrt{3}}m$
• D. $100\sqrt{3}m$

Answer 1

Answer: (C)

$\frac{100}{\sqrt{3}}m$

In, $△ABC,$

$\Rightarrow \tan 15°=\frac{50-x}{a}$

$\Rightarrow a=\frac{50-x}{\tan 15°}$

In, $△ADE,$

$\Rightarrow \tan 45°=\frac{50+x}{a}$

$\Rightarrow a=50+x$

$\Rightarrow \frac{(50-x)}{\tan 15°}=50+x$

$\Rightarrow 50-x=\tan 15°(50+x)$

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

$\Rightarrow (50-x)(\sqrt{3}+1)=(\sqrt{3}-1)(50+x)$

$\Rightarrow 50\sqrt{3}-\sqrt{3}x+50-x=50\sqrt{3}-50+\sqrt{3}x-x$

$\Rightarrow 2\sqrt{3}x=100$

$\Rightarrow x=\frac{50}{\sqrt{3}}$

$\therefore$ Height of flagpole $=2x=\frac{100}{\sqrt{3}}m$

Hence, the answer is $\frac{100}{\sqrt{3}}m.$