If $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles such that $\cos \mathrm{A}=\cos \mathrm{B}$,

Solution:

In figure $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles of $\triangle \mathrm{ABC}$.

Draw $C D \perp A B$

We are given that $\cos A=\cos B$

$\Rightarrow \frac{\mathbf{A D}}{\mathbf{A C}}=\frac{\mathbf{B D}}{\mathbf{B C}}$



$\Rightarrow \frac{\mathbf{A D}}{\mathbf{B D}}=\frac{\mathbf{A C}}{\mathbf{B C}}\left(\mathbf{F a c h}=\frac{\mathbf{C D}}{\mathbf{C D}}\right)$

$\Rightarrow \Delta \mathrm{ADC} \sim \Delta \mathrm{BDC} \quad(\mathrm{SSS}$ similarity criterion $) \Rightarrow \angle \mathrm{A}=\angle \mathrm{B}$

( $\because$ all the corresponding angles of two similar triangles are equal)