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Let H be the orthocentre of triangle ABC...

Let H be the orthocentre of triangle ABC. Then angle subtended by side BC at the centre of incircle of `Delta CHB` is

A

`(A)/(2)+90^(@)`

B

`(B+C)/(2)+90^(@)`

C

`(B-C)/(2)+90^(@)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
`angle BIC = 180^(@)-(90^(@)-C)/(2)-(90^(@)-B)/(2)`
`=(B+C)/(2)+90^(@)`
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