Question

The angle of elevation of the top of a TV tower from three points $A,B$ and $C$ in a straight line through the foot of the tower are $\alpha ,2\alpha$ and $2\alpha$ respectively. If $AB=a$, the height of the tower is

• A. $a\tan \alpha$
• B. $a\sin \alpha$
• C. $a\sin 2\alpha$
• D. $a\sin 3\alpha$

Answer 1

The correct option is C $a\sin 2\alpha$

Based on the given information, we can draw the figure shown above.

Here, $AB=a$

$E$ is the top of the TV tower

Let $BD=x$ and $ED=h$ be the height of the tower.

We know that, $\tan \theta =\frac{\text{Opposite Side}}{\text{Adjacent Side}}$

$\therefore$ In $\Delta BDE,$

$\tan 2\alpha =\frac{h}{x}$

$\Rightarrow x=\frac{h}{\tan 2\alpha }$

Also, in $\Delta ADE,$

$\tan \alpha =\frac{h}{a+x}$

$\Rightarrow \tan \alpha =\frac{h}{a+\frac{h}{\tan 2\alpha }}$

We know that, $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$

$\therefore \tan \alpha =\frac{h}{a+\frac{h(1-{\tan }^2\alpha )}{2\tan \alpha }}$

$\Rightarrow \tan \alpha =\frac{2h\tan \alpha }{2a\tan \alpha +h-h{\tan }^2\alpha }$

$\Rightarrow h(1+{\tan }^2\alpha )=2a\tan \alpha$

$\Rightarrow h=2a\sin \alpha \cos \alpha$

$\Rightarrow h=a\sin 2\alpha$ ..... $[\because \sin 2\theta =2\sin \theta \cos \theta ]$

Hence, the height of the TV tower is $a\sin 2\alpha$.