Question

The measure of angle of elevation of top of tower $75\sqrt{3}$ m high from a point at a distance of 75 m from foot of tower in a horizontal plane is :

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

Answer 1

The correct option is B ${60}^{\circ }$

Let $AB$ be the tower and $C$ be the point from which angle of elevation is observed.
Let the angle of elevation be $\theta .$

In $△ACB,$

$\Rightarrow \tan \theta =\frac{AB}{BC}=\frac{75\sqrt{3}}{75}$

$\Rightarrow \tan \theta =\sqrt{3}$

$\Rightarrow \theta ={60}^{\circ }$

Hence, the answer is ${60}^{\circ }.$