Question

One of the interior opposite angles of a triangle is ${40}^o$. If the measure of the exterior angle is ${120}^o$, find the measure of the other two angles of the triangle.

Answer 1

REF.Image

$\angle ACB+\angle ACD={180}^{\circ }$ (BCD is a straight line)

$\angle ACB=180-120={60}^{\circ }$

In $\Delta ABC$-

$\angle A+\angle B+\angle C={180}^{\circ }$

$40+\angle ABC+60={180}^{\circ }$

$\angle ABC={80}^{\circ }$