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  4. In a triangle ABC, sin text A-cos B=cosC, then angle B is

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Q. In a triangle ABC, $ sin\text{ }A-cos\,B=cosC, $ then angle B is

JamiaJamia 2009

Solution:

We have, $ sin\text{ }A-cosB=cosC $ $ \Rightarrow $ $ ~sin\text{ }A=cosB+cosC $ $ \Rightarrow $ $ 2\sin \frac{A}{2}\cos \frac{A}{2}=2\cos \left( \frac{B+C}{2} \right)\cos \left( \frac{B-C}{2} \right) $ $ \Rightarrow $ $ 2\sin \frac{A}{2}\cos \frac{A}{2}=2\sin \frac{A}{2}\cos \left( \frac{B-C}{2} \right) $ $ \Rightarrow $ $ \cos \frac{A}{2}=\cos \left( \frac{B-C}{2} \right) $ $ \left[ \because \sin \left( \frac{A}{2} \right)\ne 0 \right] $ $ \Rightarrow $ $ \frac{A}{2}=\frac{B-C}{2}\Rightarrow A=B-C $ But $ A+B+C=\pi , $ therefore $ 2B=\pi \Rightarrow B=\frac{\pi }{2} $