In ∆ABC, if ∠A + ∠B = 108° and ∠B + ∠C = 130°, find ∠A, ∠B and ∠C.
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 $\angle B+\angle C=130^{\circ}$.


$\Rightarrow \angle A+\angle B+\angle B+\angle C=(108+130)^{\circ}$

$\Rightarrow(\angle A+\angle B+\angle C)+\angle B=238^{\circ} \quad\left[\because \angle A+\angle B+\angle C=180^{\circ}\right]$

$\Rightarrow 180^{\circ}+\angle B=238^{\circ}$

$\Rightarrow \angle B=58^{\circ}$

$\therefore \angle C=130^{\circ}-\angle B$

$=(130-58)^{\circ}$

$=72^{\circ}$

$\therefore \angle A=108^{\circ}-\angle B$

$=(108-58)^{\circ}$

$=50^{\circ}$

 

 

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