Question

A tower is on a horizontal plane. The angles of elevation of top of the tower from two points on a line passing through the foot of the tower at distances 49m and 36m are ${41}^{\circ }mand{49}^{\circ }$. The height of the tower is :

• A. 40 m
• B. 42 m
• C. 44 m
• D. 46 m

Answer 1

The correct option is B

42 m

Let h be the height of the tower.


In $\Delta ABC,tan{49}^{\circ }=\frac{h}{36}\dots \left(1\right)$

$In\Delta ABD,tan{41}^{\circ }=\frac{h}{49}\Rightarrow cot({90}^{\circ }-{41}^{\circ })=\frac{h}{49}$
$\Rightarrow cot{49}^{\circ }=\frac{h}{49}\dots \left(2\right)$

Multiplying (1) and (2), we get $tan{49}^{\circ }cot{49}^{\circ }=\frac{h^2}{36\times 49}$
$\Rightarrow h^2=36\times 49\Rightarrow h=6\times 7=42m$