Question

In $△ABC$, the measure of $\angle A$ is greater than the measure of $\angle B$ by ${24}^o$. If exterior angle $\angle C$ is ${108}^o$. Find the angle of the $△ABC$.

Answer 1

Given :

$\angle A=\angle B+24$

$\angle ACD=108°$

$\therefore \angle ACB=72°$

$\therefore \angle A+\angle B+\angle C=180°$

$\angle A+24°+\angle C+72°=180°$

$2\angle B=180°-96°$

$\angle B=\frac{84°}{2}=42°$

$\angle A=66°$

$\therefore$ Angles of triangle $:66°,42°,72°$