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In an acute angled triangle ABC, AD is altitude. Circle drawn with AD as diameter, cuts the sides AB and AC at P and Q respectively. Then PQ is

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$\frac{△}{R}$

$\frac{△}{2 R}$

$\frac{2 △}{3 R}$

$\frac{△}{3 R}$

answer is A.

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Detailed Solution

In $△ P Q D$
$\begin{matrix} ∠ P D Q = 180 - [90^{\circ} - B + 90^{\circ} - C] = B + C \\ = 180^{\circ} - A \end{matrix}$

$∴ \frac{P Q}{\sin (π - A)} = 2 R^{'} = 2 \cdot$
= AD=AB sinB
$∴ P Q = c \sin B \sin A = \frac{c b \sin A}{2 R} = \frac{Δ}{R}$

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