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Find Angle A in the below figure.
"
Given:
$\angle BAC=3x^o$, $\angle ABC=5x^o$ and $\angle ACD=120^o$.
To do:
We have to find the measure of angle A.
Solution:
We know that,
The measure of each exterior angle of a triangle is equal to the sum of the opposite and non-adjacent interior angles.
Therefore,
$3x^o+5x^o=120^o$
$8x^o=120^o$
$x=\frac{120^o}{8}$
$x=15^o$
$3x^o=3(15^o)=45^o$
The measure of angle A is $45^o$.
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