Question:
In ∆ABC, if ∠A + ∠B = 108° and ∠B + ∠C = 130°, find ∠A, ∠B and ∠C.
Solution:
Let $\angle A+\angle B=108^{\circ}$ and $\angle B+\angle C=130^{\circ}$.
$\Rightarrow \angle A+\angle B+\angle B+\angle C=(108+130)^{\circ}$
$\Rightarrow(\angle A+\angle B+\angle C)+\angle B=238^{\circ} \quad\left[\because \angle A+\angle B+\angle C=180^{\circ}\right]$
$\Rightarrow 180^{\circ}+\angle B=238^{\circ}$
$\Rightarrow \angle B=58^{\circ}$
$\therefore \angle C=130^{\circ}-\angle B$
$=(130-58)^{\circ}$
$=72^{\circ}$
$\therefore \angle A=108^{\circ}-\angle B$
$=(108-58)^{\circ}$
$=50^{\circ}$