Question

In a triangle ABC, $\sin A-\cos B=\cos C$, then what is B equal to?

• A. $\pi$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (C)

$\frac{\pi }{2}$

In a triangle $ABC,\sin A-\cos B=\cos C$

$\sin A=\cos B+\cos C$

$2\sin \frac{A}{2}\cos \frac{A}{2}=2\cos \frac{B+C}{2}\cos \frac{B-C}{2}$

$\sin \frac{B+C}{2}\cos \frac{B-C}{2}$

In $\Delta ABC$

$A+B+C=\pi$

$\frac{A}{2}=\frac{\pi -(B+C)}{2}$

$\frac{(B+C)}{2}=\frac{\pi }{2}-\frac{(B-C)}{2}$

$\frac{(B+C)}{2}=\frac{\pi }{2}-\frac{B}{2}+\frac{C}{2}$

$B=\frac{\pi }{2}$