Question

In $\Delta ABC,\angle A+\angle B={65}^0$ and $\angle B+\angle C={140}^0$, find the value of $\angle B$ and $\angle C$.

Answer 1

We know that in a $△ABC=\angle A+\angle B+\angle C={180}^{\circ }$ ......$\left(1\right)$

Given:$\angle A+\angle B={65}^{\circ }$

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

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

$\Rightarrow \angle A+2\angle B+\angle C={205}^{\circ }$ ......$\left(2\right)$

Eqn$\left(2\right)-$Eqn$\left(1\right)$ we get

$\angle A+2\angle B+\angle C-\angle A-\angle B-\angle C={205}^{\circ }-{180}^{\circ }={25}^{\circ }$

$\Rightarrow \angle B={25}^{\circ }$

Substituting $\angle B={25}^{\circ }$ in $\angle A+\angle B={65}^{\circ }$

we get

$\angle A={65}^{\circ }-{25}^{\circ }={40}^{\circ }$

Substituting $\angle B={25}^{\circ }$ in $\angle B+\angle C={140}^{\circ }$

$\angle C={140}^{\circ }-\angle B={140}^{\circ }-{25}^{\circ }={125}^{\circ }$

$\therefore \angle B={25}^{\circ },\angle C={125}^{\circ }$