Q1. [1]
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) $60°$
Previously asked in CBSE board exam
2026 30/1/1 Q11
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding rag
Model Answer
(d) 60°
Using $\tan\theta = \dfrac{\text{height}}{\text{distance}} = \dfrac{30}{10\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3}$, so $\theta = 60°$.
Explanation
Set up a right triangle: tower height = 30 m (perpendicular), base distance = 10√3 m. Apply tan(angle of elevation) = opposite/adjacent = 30/10√3 = √3, which corresponds to 60°. This is a direct application of Example 1-style problems from Chapter 9.
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