Question

In $△ABC$, if $\cos A=\sin B-\cos C$, then show that it is a right angled triangle.

Answer 1

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

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

$2\cos \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)=\sin \frac{B}{2}\cos \frac{B}{2}$

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

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

$\cos \left(\frac{A+C}{2}\right)=\sin \frac{A}{2}$

$2\sin /\frac{B}{2}\cos \left(\frac{A-C}{2}\right)=2\sin /\frac{B}{2}\cos \frac{B}{2}$

$\frac{A-C}{2}=\frac{B}{2}$

$A=B+C$

$A+B+C=\pi$

$A+A=\pi \Rightarrow 2A=\pi \Rightarrow A=\pi /2$