Question

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`

Answer 1

Correct Answer - 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`