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In triangle ABC, $∠ A = 30^{\circ}, B C = 2 + \sqrt{5},$ , then the distance of the vertex A from the.orthocentre of the triangle is

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$(2 + \sqrt{5}) \sqrt{3}$

$\frac{\sqrt{3} + 1}{2 \sqrt{2}}$

$\frac{1}{2}$

answer is B.

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Detailed Solution

$R = \frac{a}{2 \sin A} = \frac{2 + \sqrt{5}}{2 \sin 30^{\circ}} = \frac{2 + \sqrt{5}}{2 \times \frac{1}{2}} = (2 + \sqrt{5})$
$Now, A H = 2 R \cos A = 2 (2 + \sqrt{5}) \cos 30^{\circ} = (2 + \sqrt{5}) \sqrt{3}$

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