Question

In $∆ABC,\mathrm{if}\cos A=\sin B-\cos C$, show that it is a right-angled triangle.

Answer 1

Let ABC be any triangle.

$$\begin{gathered} \mathrm{Given},\cos A=\sin B-\cos C \\ \Rightarrow \cos A+\cos C=\sin B \\ \Rightarrow 2\cos \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)=2\sin \frac{B}{2}\cos \frac{B}{2} \\ \Rightarrow \cos \left(\frac{\mathrm{\pi }}{2}-\frac{B}{2}\right)\cos \frac{A-C}{2}=\sin \frac{B}{2}\cos \frac{B}{2}\left(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{\pi }\right) \\ \Rightarrow \sin \frac{B}{2}\cos \left(\frac{A-C}{2}\right)=\sin \frac{B}{2}\cos \frac{B}{2} \\ \Rightarrow \cos \left(\frac{A-C}{2}\right)=\cos \frac{B}{2} \\ \Rightarrow \frac{A-C}{2}=\frac{B}{2} \\ \Rightarrow A-C=B...\left(1\right) \\ \mathrm{Now}, \\ \because A+B+C=180° \\ \therefore A+A-C+C=180° \\ \Rightarrow 2A=180° \\ \Rightarrow A=90° \\ \mathrm{So},∆ABC\mathrm{is}\mathrm{a}\mathrm{triangle}\mathrm{which}\mathrm{is}\mathrm{right}\mathrm{angled}\mathrm{at}A\mathit{.} \\ \mathrm{Hence},\mathit{}∆\mathrm{ABC}\mathit{}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angled}\mathrm{triangle}. \end{gathered}$$