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In a triangle ABC, sin A- cos B = Cos C...

In a triangle `ABC, sin A- cos B = Cos C`, then angle `B` is

A

`pi//2`

B

`pi//3`

C

`pi//4`

D

`pi//6`

Text Solution

Verified by Experts

The correct Answer is:
A
We have, `sin A-cosB=cosC`
`sinA=cosB+cosC`
`rArr 2 sin(A)/(2)cos(A)/(2)=2cos((B+C)/(2))cos((B-C)/(2))`
`rArr 2sin(A)/(2)cos(A)/(2)=2cos((pi-A)/(2))cos((B-C)/(2))because A+B+C=pi`
`rArr2sin(A)/(2)cos(A)/(2)=2sin(A)/(2)cos((B-C)/(2))`
`rArrcos(A)/(2)=cos(B-C)/(2)` or `A=B-C: "but" A+B+C=pi`
therefore `2B=pirArrB=pi//2`
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