In triangle ABC, $\angle A=80^o, \angle B=30^o$. Which side of the triangle is the smallest?
Given:
In triangle ABC, $\angle A=80^o, \angle B=30^o$.
To do:
We have to find the smallest side of the triangle.
Solution:
We know that,
In any triangle side opposite to the smallest angle is the shortest side.
Therefore,
$\angle A+\angle B+\angle C=180^o$
$80^o+30^o+\angle C=180^o$
$\angle C=180^o-110^o=70^o$
The smallest angle in the given triangle is $\angle B$.
This implies,
The side opposite to $\angle B$, that is, $AC$ is the shortest side.
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