Q1. [1]
From a point on the ground, which is 60 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is found to be 45°. 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
2026 30/2/1 Q12
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding rag
Model Answer
Option C: 60
Using $\tan 45° = \dfrac{\text{height}}{\text{distance}} = \dfrac{h}{60}$, and since $\tan 45° = 1$, we get $h = 60$ m.
Explanation
Apply $\tan(\text{angle of elevation}) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{h}{60}$. Since $\tan 45° = 1$, height = 60 m directly. No surds involved — a common trap is confusing this with 30° or 60° cases.
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