Question

In a triangle ABC, $sinA-cosB=cosC,$ then angle B is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (A)

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