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In a DeltaABC, the bisectors of angleBan...

In a `DeltaABC,` the bisectors of `angleBandangleC` intersect each other point O. Then`angleBOC` is ?

A

`angleBOC=180^(@)+(1)/(2)angleA`.

B

`angleBOC=90^(@)+(1)/(2)angleA`.

C

`angleBOC=180^(@)-(1)/(2)angleA`.

D

`angleBOC=90^(@)-(1)/(2)angleA`.

Text Solution

Verified by Experts

The correct Answer is:
B
A `DeltaABC` in which BO and CO are the bisectors of `angleBandangleC` respectively.
TO PROVE `angleBOC=(90^(@)+(1)/(2)angleA)`.
We know that the sum of the angles of a triangle is `180^(@)`.
`:.angleA+angleB+angleC=180^(@)`
`implies(1)/(2)angleA+(1)/(2)angleB+(1)/(2)angleC=90^(@)`
`implies(1)/(2)angleA+angleOBC+angleOCB=90^(@)`
`impliesangleOBC+angleOCB=(90^(@)-(1)/(2)angleA)" "....(i)`
Nwo, in `DeltaOBC` we have
`angleOBC+angleOCB+angleBOC=180^(@)`
[sum of the angles of a triangle]
`implies(90^(@)-(1)/(2)angleA)+angleBOC=180^(@)" "["using (i)"]`
`impliesangleBOC=180^(@)-(90^(@)-(1)/(2)angleA)=(90^(@)+(1)/(2)angleA)`.
Hence, `angleBOC=(90^(@)+(1)/(2)angleA)`.
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