A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{6}\]
Correct Answer: A
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 \[2B=\pi \Rightarrow B=\frac{\pi }{2}\]
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