If $\angle A$ and $\angle B$ are acute angles such that $cos\ A = cos\ B$, then show that $∠A = ∠B$.

Given:

\( \angle A \) and \( \angle B \) are acute angles such that \( \cos A=\cos B \).

To do:

We have to show that \( \angle A=\angle B \).

Solution:  

Let, in a triangle $ABC$ right angled at $C$, $cos\ A = cos\ B$.

We know that,

In a right-angled triangle $ABC$ with a right angle at $C$,

By trigonometric ratios definitions,

$cos\ A=\frac{Adjacent}{Hypotenuse}=\frac{AC}{AB}$

$cos\ B=\frac{Adjacent}{Hypotenuse}=\frac{BC}{AB}$

This implies,

$\cos A=\cos B$

$\Rightarrow \frac{AC}{AB}=\frac{BC}{AB}$

 $\Rightarrow AC=BC$

We know that,

Angles opposite to equal sides are equal in a triangle.

Therefore,

$\angle A=\angle B$

Hence proved.