Question

Question 124

In $\Delta ABC$, if $\angle A=\angle C$ and exterior $\angle ABX={140}^{\circ }$, then find the angles of the triangle.

Answer 1

Given, $\angle A=\angle C$ and exterior $\angle ABX={140}^{\circ }$
Let $\angle A=\angle C=x$
According to the exterior angle property,

Exterior $\angle B=$ Interior $\angle A$ + Interior $\angle C$
$\Rightarrow {140}^{\circ }=x+x\Rightarrow {140}^{\circ }=2x\Rightarrow x=\frac{{140}^{\circ }}{2}={70}^{\circ }$
So, $\angle A=\angle C={70}^{\circ }$
Now, $\angle A+\angle B+\angle C={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow {70}^{\circ }+\angle B+{70}^{\circ }={180}^{\circ }\Rightarrow \angle B+{140}^{\circ }={180}^{\circ }\Rightarrow \angle B={180}^{\circ }-{140}^{\circ }\Rightarrow \angle B={40}^{\circ }$
Hence, all the angles of the triangle are ${70}^{\circ },{40}^{\circ }$ and ${70}^{\circ }$