In $a \triangle A B C, \angle A+\angle B=65^{\circ}$ and $\angle B+\angle C=140^{\circ}$. Find the angles.
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In $a \triangle A B C, \angle A+\angle B=65^{\circ}$ and $\angle B+\angle C=140^{\circ}$. Find the angles.
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Solution-

$\angle A+\angle B=65^{\circ}$ and $\angle B+\angle C=140^{\circ}$

$\Rightarrow \angle A+\angle B+\angle B+\angle C=\left(65^{\circ}+140^{\circ}\right)$

$\Rightarrow \quad(\angle A+\angle B+\angle C)+\angle B=205^{\circ}$

$\Rightarrow 180^{\circ}+\angle B=205^{\circ} \quad\left[\because \angle A+\angle B+\angle C=180^{\circ}\right]$

$\Rightarrow \angle B=\left(205^{\circ}-180^{\circ}\right)=25^{\circ}$.

$\therefore \angle C=\left(140^{\circ}-\angle B\right)=\left(140^{\circ}-25^{\circ}\right)=115^{\circ}$

and $\angle A=\left(65^{\circ}-\angle B\right)=\left(65^{\circ}-25^{\circ}\right)=40^{\circ}$.

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