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 $3 \alpha$ respectively. If $A B=a$, then height of the tower is

• A

  $a \tan \alpha$
• B

  $a \sin \alpha$
• Correct answer
  C

  $a \sin 2 \alpha$
• D

  $a \sin 3 \alpha$

Solution

In $\triangle \mathrm{ABE}$,

In $\triangle \mathrm{BCE}$, using sine rule, $$ \begin{aligned} \frac{\mathrm{a}}{\sin \left(180^{\circ}-3 \alpha\right)} &=\frac{\mathrm{CE}}{\sin 2 \alpha} \\ \Rightarrow \quad \mathrm{CE} &=\frac{\mathrm{a} \sin 2 \alpha}{\sin 3 \alpha} \end{aligned} $$ Now, in $\triangle \mathrm{DCE}$, $$ \begin{aligned} &\sin 3 \alpha=\frac{h}{C E} \\ &\Rightarrow \quad \sin 3 \alpha=\frac{h}{a \sin 2 \alpha / \sin 3 \alpha} \quad \text { [from Eq. (i)] } \\ &\Rightarrow \quad \mathrm{h}=\mathrm{a} \sin 2 \alpha \end{aligned} $$

Hence, the correct answer is option: 'C'.