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