In $\triangle ABC$, if $\angle A = 40^o$ and $\angle B = 60^o$. Determine the longest and shortest sides of the triangle.
Given:
In $\triangle ABC$, $\angle A = 40^o$ and $\angle B = 60^o$.
To do:
We have to determine the longest and shortest sides of the triangle.
Solution:
We know that,
Sum of the angles in a triangle is $180^o$.
Therefore,
$\angle A +\angle B +\angle C =180^o$
$40^o+60^o+\angle C=180^o$
$\angle C=180^o-100^o$
$\angle C=80^o$
$\angle C$ is the largest angle. This implies side $AB$ which is opposite to the greatest angle is the longest side.
Similarly,
$\angle A$ is the smallest angle. This implies side $BC$ which is opposite to the smallest angle is the shortest side.
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