MCQ
In a $\triangle ABC , \angle C =3 \angle B=2(\angle A+\angle B )$, then $\angle B =$ ?
- A$20^{\circ}$
- ✓$40^{\circ}$
- C$60^{\circ}$
- D$80^{\circ}$
✓
Answer
Correct option: B.
$40^{\circ}$
Let $C =3 B=2(A+ B )= x ^{\circ}$
Then, $c = x ^{\circ}, B =\left(\frac{ x }{3}\right)^{\circ}$ and $( A + B )=\left(\frac{ x }{2}\right)^{\circ}$
$(A+B)+C=180^{\circ} $
$\Rightarrow \frac{x}{2}+x=180^{\circ}$
$\Rightarrow 3 x=360 $
$\Rightarrow x=\frac{360}{3}=120^{\circ}$
$\therefore \angle B=\left(\frac{120}{3}\right)^{\circ}$
$=40^{\circ}$
Then, $c = x ^{\circ}, B =\left(\frac{ x }{3}\right)^{\circ}$ and $( A + B )=\left(\frac{ x }{2}\right)^{\circ}$
$(A+B)+C=180^{\circ} $
$\Rightarrow \frac{x}{2}+x=180^{\circ}$
$\Rightarrow 3 x=360 $
$\Rightarrow x=\frac{360}{3}=120^{\circ}$
$\therefore \angle B=\left(\frac{120}{3}\right)^{\circ}$
$=40^{\circ}$
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