Question

The angle elevation of the top of a tower from two points situated at distances $36m$ and $64m$ from its base and in the same straight line with it are complementary. What is the height of the tower?

• A. 50 m
• B. 48 m
• C. 25 m
• D. 24 m

Answer 1

Answer: (B)

48 m

Let $AB$ be the tower.

Angle of elevation at a distance of $36$m at point $C=\theta$

Angle of elevation at a distance of $64$m at the point $D={90}^{\circ }-\theta$

In $△ABC,\cot \theta =\frac{36}{AB}....\left(1\right)$

In $△ADB,\tan ({90}^{\circ }-\theta )=\frac{AB}{64}$

$\Rightarrow \cot \theta =\frac{AB}{64}.....\left(2\right)$

From $\left(1\right)\&\left(2\right)$, we have $\cot \theta =\frac{36}{AB}=\frac{AB}{64}$

$\Rightarrow \frac{36}{AB}=\frac{AB}{64}$

$\Rightarrow {\left(AB\right)}^2=36\times 64$

$AB=\sqrt{36\times 64}=6\times 8=48m$