1.

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

Answer»

`(A)/(2)+90^(@)`
`(B+C)/(2)+90^(@)`
`(B-C)/(2)+90^(@)`
none of these

Solution :`ANGLE BIC = 180^(@)-(90^(@)-C)/(2)-(90^(@)-B)/(2)`
`=(B+C)/(2)+90^(@)`


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