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

Step 1: Understanding the Concept:
The problem describes a right-angled triangle formed by the height of the tower, the distance of the car from the base, and the line of sight (hypotenuse). We use the tangent ratio to find the angle of elevation.
Step 2: Key Formula or Approach:
\[ \tan \theta = \frac{\text{Perpendicular (Height)}}{\text{Base (Distance)}} \]
Step 3: Detailed Explanation:
1. Let the height of the tower be \( h = 30 \) m. 2. Let the distance of the car from the base be \( b = 10\sqrt{3} \) m. 3. Let the angle of elevation be \( \theta \). 4. Applying the formula: \[ \tan \theta = \frac{30}{10\sqrt{3}} \] \[ \tan \theta = \frac{3}{\sqrt{3}} \] \[ \tan \theta = \sqrt{3} \] 5. We know that \( \tan 60^\circ = \sqrt{3} \), so \( \theta = 60^\circ \).
Step 4: Final Answer:
The angle of elevation is 60°.