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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

Text Solution

Verified by Experts

The correct Answer is:
`(2 + sqrt5) sqrt3`
Distance of vertex a from orthocenter H given by is
`AH = 2 R cos A`
we have,
`R = (a)/(2 sin A) = (2 + sqrt5)/(2 sin 30^(@)) = (2 + sqrt5)/(2 xx (1)/(2)) = (2 + sqrt5)`
`:. AH = 2 (2 + sqrt5) cos 30^(2)`
`= (2 + sqrt5) sqrt3`
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