Q1. [1] § 9.1 Heights and Distances § 9.2 Summary
From a point on the ground, which is 30 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is found to be 60°. The height (in metres) of the tower is :
- A $10\sqrt{3}$
- B $30\sqrt{3}$
- C $60$
- D $30$
Previously asked in CBSE board exam
2024 30/3/1 Q7
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Option B: $30\sqrt{3}$ m
Using $\tan 60° = \dfrac{h}{30}$, we get $\sqrt{3} = \dfrac{h}{30}$, so $h = 30\sqrt{3}$ m.
Explanation
This directly applies $\tan(\text{angle of elevation}) = \dfrac{\text{height}}{\text{base distance}}$. The base is 30 m and angle is 60°, so multiply 30 by $\tan 60° = \sqrt{3}$. This mirrors Example 1 of Chapter 9 (where base = 15 m gave $15\sqrt{3}$). Don't confuse with Exercise Q4 (which has the same base but angle 30°, giving a different answer).
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