Question

A car is moving away from the base of a 30 m high tower. The angle of elevation of the top of the tower from the car at an instant, when the car is \(10\sqrt{3}\) m away from the base of the tower, is :

• A. 30°
• B. 45°
• C. 90°
• D. 60°

Hint:

To simplify \(\frac{3}{\sqrt{3}}\), rationalize the denominator: \(\frac{3 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}\).

Answer 1

The Correct Option is D

To solve the problem, we need to determine the angle of elevation of the top of the tower from the car when the car is at a specific distance from the base of the tower.

Given:

• Height of the tower (\(h\)) = 30 m
• Distance of the car from the base of the tower (\(d\)) = \(10\sqrt{3}\) m

We need to find the angle of elevation (\(\theta\)) from the car to the top of the tower. The angle of elevation can be found using the tangent function in trigonometry, which is given by:

\[\tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}}\]

Here, the opposite side is the height of the tower, and the adjacent side is the distance of the car from the base of the tower. Substituting the given values, we have:

\[\tan(\theta) = \frac{30}{10\sqrt{3}}\]

Simplifying, we get:

\[\tan(\theta) = \frac{30}{10\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}\]

We know that:

\[\tan(60^\circ) = \sqrt{3}\]

Therefore, the angle of elevation \(\theta\) is:

60°

Thus, the correct answer is 60°.

This conclusion is based on the basic trigonometric identity where the tangent of 60° equals \(\sqrt{3}\), and understanding of the relationship between the elements in a right triangle.