In a $\triangle ABC$, if $\angle A = 55^o, \angle B = 40^o$, find $\angle C$.
Given:
In a $\triangle ABC$, $\angle A = 55^o, \angle B = 40^o$.
To do:
We have to find $\angle C$.
Solution:
We know that,
Sum of the angles in a triangle is $180^o$.
Therefore,
$\angle A + \angle B + \angle C = 180^o$
$55^o+40^o+\angle C=180^o$
$\angle C=180^o-95^o$
$\angle C=85^o$
Hence, $\angle C=85^o$.
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