Question

The height of a tower is $100\sqrt{3}$. The angle of elevation of its top from a point $100m$ away from its foot(base) is ______.

• A. 30${}^o$
• B. 45${}^o$
• C. 60${}^o$
• D. 90${}^o$

Answer 1

The correct option is C 60${}^o$

$AB=$ tower, $O$ is the point of observation.

Given that height of the tower is $100$$\sqrt3$

$\therefore$ $AB=100$$\sqrt{3}$, $OA=100m$

$\therefore$ $\tan$ $\theta$ = $\frac{AB}{OA}=\frac{100\sqrt{3}}{100}$ $=$ $\sqrt{3}$

$\therefore \theta$ $=60$ ${}^o$

$\therefore$ Angle of elevation is ${60}^o$.

So, option C is correct.