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In any triangle ABC, sin A-cos B=cosC, t...

In any triangle ABC, `sin A-cos B=cosC`, 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, `sinA-cosB=cosC`
`sinA=cosB+cosC`
`rArr 2sinA/2cosA/2=2cos(B+C)/2cos(B-C)/2`
`rArr 2sinA/c cosA/2=2cos(pi-A)/2 cos(B-C)/(2)` `therefore A+B+C=pi`
`rArr 2sinA/2 cosA/2=2sinA/2cos(B-C)/(2)`
`rArr cosA/2=cos(B-C)/(2)` or `A=B-C`, But A+B+C=`pi`
Therefore `2B=pi rArr B=pi/2`
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