Q1. [1] § 9.1 Heights and Distances § 9.2 Summary
A wire is attached from a point $A$ on the ground to the top of a pole $BC$, making an angle of elevation as $60°$. If $AB = 5\sqrt{3}$ m, then length of the wire is
- (A) $10$ m
- (B) $10\sqrt{3}$ m
- (C) $15$ m
- (D) $\frac{5}{\sqrt{3}}$ m
Previously asked in CBSE board exam
2026 30/5/1 Q17
Generated by claude-sonnet-4-6 · 2026-06-15 10:37 · grounding rag
Model Answer
(A) 10 m
In right △ABC, AB = 5√3 m (base), angle of elevation = 60°.
cos 60° = AB/AC → 1/2 = 5√3/AC → AC = 10 m.
Explanation
The wire is the hypotenuse (AC). Use cos 60° = adjacent/hypotenuse = AB/AC. Since cos 60° = 1/2, AC = 2 × 5√3 is wrong — note AB here is the horizontal distance, so AC = AB/cos60° = 5√3 ÷ (1/2) = 10 m. Always identify which side is the hypotenuse (the wire/rope) and which trig ratio links the given side and required side.
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