Question

The angle of elevation of the top of a T.V. Tower from three points A, B, C in a st. line through the foot of the tower are $\alpha ,2\alpha ,3\alpha$ 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

Answer: (C)

Let OP be a vertical tower. The elevation of top P from A, B, C are respectively $\alpha$ , 2$\alpha$ , 3$\alpha$ respectively.
From, rt. $\Delta d$ = OBP , $\frac{OP}{BP}$ = sin $2\alpha$
$\therefore$ OP = BP sin $2\alpha =a$ sin $2\alpha$
Thus, height of the tower = $a$ sin $2\alpha$.