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Question:

A vertical tower OP stands at the centre O of a square ABCD. Let h and b denote the lengths OP and AB respectively. Suppose ∠APB = 60°, then the relationship between h and b can be expressed as

Explanation:

OP = h and AB = b

Now, OA = $\frac{AC}{2}=\frac{b\sqrt{2}}{2}=\frac{b}{\sqrt{2}}$

OP is a perpendicular tower at the centre O of the square.

In ∆PAB, PA = PB

∴ ∠PAB = ∠PBA = ∠APB = 60°

∴ ∆PAB is an equilateral triangle.

∴ AP = b

In the right-angled ∆AOP, we have,

AP² = OP² + OA²

∴ b² = h² + $\frac{b^2}{2}$

∴ 2h² = b²

Hence, option (b).