A tower stands vertically on the ground. From a point on the ground, 30 m away fromthe foot of the tower, the angle of elevation of the top of the tower is 30°. What is theheight (in metres) of the tower?

Correct option is D

Given:

The distance from the point on the ground to the foot of the tower = 30 m

The angle of elevation of the top of the tower = 30°

Formula Used:

We can use the tangent function in trigonometry to solve this problem.

The tangent of the angle of elevation is the ratio of the height of the tower (h) to the horizontal distance (d) from the point to the foot of the tower:

​$$\tan(\theta) = \frac{\text{height of the tower}}{\text{distance from the foot of the tower}}\\\ \\\tan(\theta) = \frac{h}{d}$$​​

Solution: 

1. Substitute the given values:

​$$\tan(30^\circ) = \frac{h}{30} \\\ \\\tan(30^\circ) = \frac{1}{\sqrt{3}}\\\ \\\frac{1}{\sqrt{3}} = \frac{h}{30}\\\ \\h = 30 \times \left( \frac{1}{\sqrt{3}} \right) = \frac{30}{\sqrt{3}} = 30 \times \left( \frac{\sqrt{3}}{3} \right)\\\ \\h = 10\sqrt{3}$$​​

Therefore, the height of the tower is approximately $$10\sqrt{3}$$​ meters