Question

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

Answer 1

Dear student,

$$\begin{gathered} Given:\cos A=\sin B-\cos C \\ \Rightarrow \cos A+\cos C=\sin B \\ \Rightarrow 2\cos \left(\frac{A+C}{2}\right).\cos \left(\frac{A-C}{2}\right)=\sin \left[180-(A+C)\right] \\ \Rightarrow 2\cos \left(\frac{A+C}{2}\right).\cos \left(\frac{A-C}{2}\right)=\sin (A+C) \\ \Rightarrow 2\cos \left(\frac{A+C}{2}\right).\cos \left(\frac{A-C}{2}\right)=2\sin \left(\frac{A+C}{2}\right)\cos \left(\frac{A+C}{2}\right) \\ \Rightarrow \cos \left(\frac{A-C}{2}\right)=\sin \left(\frac{A+C}{2}\right) \\ \Rightarrow \cos \left(\frac{A-C}{2}\right)=\cos \left[90-\left(\frac{A+C}{2}\right)\right] \\ \Rightarrow \left(\frac{A-C}{2}\right)=90-\left(\frac{A+C}{2}\right) \\ \Rightarrow A-C=180-(A+C) \\ \Rightarrow A-C=180-A-C \\ \Rightarrow 2A=180 \\ \Rightarrow A=90 \\ Thus,∆ABCisrightangledtriangle. \end{gathered}$$

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