Question

In $\Delta ABC$, if a $\cos A=b\mathrm{cosB}$, then prove that the triangle is either a right-angled or an isosceles triangle.

Answer 1

We have, $a\cos A=b\cos B$

$\Rightarrow k\sin A\cos A=k\sin B\cos B$ by sine formula

or $2\sin A\cos A=2\sin B\cos B$

or $\sin 2A=\sin 2B$

or $\sin 2A-\sin 2B=0$

or $2\cos (A+B)\sin (A-B)=0$

$\therefore$ either $\cos (A+B)=0$

$\Rightarrow A+B={90}^{\circ }$

So that $\angle C={90}^{\circ }$

$\Rightarrow △ABC$ is right-angled at $C$ or $\sin (A-B)=0$

i.e.,$A-B=0$

so that $A=B\Rightarrow △ABC$ is isosceles.