Question

In $\Delta ABC,if\angle A={40}^o$ and $\angle B={80}^o$,then find smallest side of triangle.

Answer 1

$\angle A={40}^{\circ }$, $\angle B={80}^{\circ }$
Sum of angles of triangle = 180
$\angle A+\angle B+\angle C=180$
$40+80+\angle C=180$
$\angle C={60}^{\circ }$
In a triangle the smallest side is opposite to the smallest angle. Since, $\angle A$ is the smallest angle. hence, BC is the smallest side.