Question

One of the exterior angles of a triangle is ${80}^o$, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?

Answer 1

Given: $\angle BCO={80}^{\circ }$ and $\angle BAC=\angle ABC=x^{\circ }\left[\text{say}\right]$

To find: $\angle BAC,\angle ABC$

Solution: By exterior angle sum property of triangle, we have

$\angle BCO=\angle BAC+\angle ABC$

$\Rightarrow {80}^{\circ }=x^{\circ }+x^{\circ }$

$\Rightarrow {80}^{\circ }=2x^{\circ }$

$\Rightarrow x^{\circ }=(\frac{80}{2}{)}^{\circ }$

$\Rightarrow x^{\circ }={40}^{\circ }$

$\therefore \angle BAC=\angle ABC={40}^{\circ }$