Question

If $\frac{\cos A+2\cos C}{\cos A+2\cos B}=\frac{\sin B}{\sin C}$ the the triangle
$ABC$ is either isosceles or right angled.

Answer 1

$\cos A(\sin B-\sin C)+(\sin 2B-\sin 2C)=0$
or $\cos A(\sin B-\sin C)+2\sin (B-C)\cos (B+C)=0$
or $\cos A(\sin B-\sin C)-2\cos A\sin (B-C)=0$
$\therefore$ either $\cos A=0$ $\Rightarrow A={90}^o$
or $(\sin B-\sin C)-2(\sin B\cos C-\cos B\sin C)=0$
or $a(b-c)-2(b^2-c^2)=0$
$\Rightarrow (b-c)-[a-2(b+c\left)\right]=0$
$\therefore b-c=0$ $b=c$ $\therefore$ Isosceles.