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In a triangle ABC, angle A = 30^(2), BC ...

In a triangle ABC, `angle A = 30^(2), BC = 2 + sqrt5`, then find the distance of the vertex A from the orthocenter

A

1

B

`(2+sqrt(5)) sqrt(3)`

C

`(sqrt(3)+1)/(2 sqrt(2))`

D

`(1)/(2)`

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The correct Answer is:
B
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