Question

In a$∆ABC$, if $\sin A-\cos B=\cos C$, then $\angle B$ is equal to

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

Answer 1

The correct option is A

$\frac{\pi }{2}$

Explanation for correct option

Given triangle $ABC$ in which $\sin \left(A\right)-\sin \left(B\right)=\cos \left(C\right)$

$A,B$ and $C$ are the angles of a triangle, hence,

$A+B+C=\pi \left(1\right)$

Now,

$$\begin{gathered} \Rightarrow \sin \left(A\right)-\cos \left(B\right)=\cos \left(C\right) \\ \Rightarrow \sin \left(A\right)=\cos \left(B\right)+\cos \left(C\right) \\ \Rightarrow 2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)=2\cos \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)\left[\because \cos \left(A\right)+\cos \left(B\right)=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\right] \\ \Rightarrow \sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)=\cos \left(\frac{\pi -A}{2}\right)\cos \left(\frac{B-C}{2}\right)\left[\because A+B+C=\pi \right] \\ \Rightarrow \sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)=\sin \left(\frac{A}{2}\right)\cos \left(\frac{B-C}{2}\right) \\ \Rightarrow \cos \left(\frac{A}{2}\right)=\cos \left(\frac{B-C}{2}\right) \end{gathered}$$

Therefore we can equate,

$$\begin{gathered} \frac{A}{2}=\frac{B-C}{2} \\ \Rightarrow A=B-C \\ \Rightarrow \pi -B-C=B-C\left[\because A=\pi -B-C\right] \\ \Rightarrow 2B=\pi \\ \Rightarrow B=\frac{\pi }{2} \end{gathered}$$

Therefore $\angle B$ is equal to $\frac{\pi }{2}$.

Hence, the correct answer is option (A).