If $\text{ a }\cos A=b \cos B$ then the triangle, is

We have,

$\begin{array}{l}\Rightarrow a\left(\frac{b^2+c^2-a^2}{2bc}\right)=b\left(\frac{c^2+a^2-b^2}{2ac}\right)\\ \Rightarrow a^2{b}^2+a^2{c}^2-a^4=b^2{c}^2+b^2{a}^2-b^4\\ \Rightarrow c^2\left(a^2-b^2\right)-\left(a^4-b^4\right)=0\\ \Rightarrow a=b\text{ or, }{c}^2=a^2+b^2\end{array}$

$\Rightarrow ∆ ABC$ is isosceles or right angled.