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 $\alpha ,2\alpha and3\alpha ,$ respectively. If$AB=a$, then 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

$a\sin 2\alpha$

Explanation for the correct option:

Step 1: Drawing the diagram

Let $h$ be the height of the tower $ED$.

Given,$AB=a$

Let $BD=y$

$\therefore AD=a+y$

Step 2: Find the height of the tower:

In $∆ADE$,

$$\begin{gathered} \tan \alpha =\frac{ED}{AD} \\ \Rightarrow \tan \alpha =\frac{h}{a+y}...\left(1\right) \end{gathered}$$

In $∆BDE$

$$\begin{gathered} \tan 2\alpha =\frac{ED}{BD} \\ \Rightarrow \tan 2\alpha =\frac{h}{y}...\left(2\right) \end{gathered}$$

From equation (1) and (2)

$$\begin{gathered} \tan \alpha (a+y)=y\tan 2\alpha \\ \Rightarrow \tan \alpha (a+y)=y\frac{2\tan \alpha }{1-{\tan }^2\alpha }\mathbf{}\left(\because tan2\alpha =\frac{2tan\alpha }{1-ta{n}^2\alpha }\right) \\ \Rightarrow 2y=(a+y)-(a+y){\tan }^2\alpha \\ \Rightarrow y+y{\tan }^2\alpha =a(1-{\tan }^2\alpha ) \\ \Rightarrow y=a.\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha } \\ \Rightarrow h=\left(a.\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }+a\right)\tan \alpha \\ \Rightarrow h=\frac{2a\tan \alpha }{se{c}^2\alpha } \\ \Rightarrow h=2a\sin \alpha \cos \alpha \\ \Rightarrow h=a\sin 2\alpha \left(\because 2sin\alpha cos\alpha =sin2\alpha \right) \end{gathered}$$

Hence, option (C) is the correct answer.