Question

In $△ABC,\angle A={45}^o$ and $\angle B={65}^o$. Name the side of the triangle which is
shortest

Answer 1

$\angle A=45°$

$\angle B=65°$

Now, in $△ABC$,

$\angle A+\angle B+\angle C=180°\left(\text{Sum of interior angles}\right)$

$\Rightarrow \angle C=180°-45°-65°=180°-110°=70°$

As we know that the side opposite the smallest angle is shortest.

Hence side $BC$ is shortest.