MCQ
In a $\ce{\triangle ABC}$, if $\angle A-\angle B=42^{\circ}$ and $\angle B-\angle C=21^{\circ}$ then $\angle B= ?$
- A$95^{\circ}$
- B$63^{\circ}$
- ✓$53^{\circ}$
- D$32^{\circ}$
✓
Answer
Correct option: C.
$53^{\circ}$
Let $\angle A-\angle B=42^{\circ}\ldots(i)$ and
$\angle B-\angle C=21^{\circ}\ldots(ii)$
Adding $(i)$ and $(ii),$ we get
$\angle A-\angle C=63^{\circ} \ldots(iii)$
$\angle B=\angle A-42^{\circ} \ldots \ldots .[$ using $(i)]$
$\angle C=\angle A-63^{\circ} [$Using $(iii)]$
$\therefore \angle A+\angle B+\angle C=180^{\circ} [$ Sum of the angles of a triangle$] $
$\Rightarrow \angle A+\angle A-42^{\circ}+\angle A-63^{\circ}=180^{\circ}$
$\Rightarrow 3 \angle A-105^{\circ}=180^{\circ}$
$\Rightarrow 3 \angle A=285^{\circ}$
$\therefore \angle B=(95-42)^{\circ}$
$\Rightarrow \angle B=53^{\circ}$
$\angle B-\angle C=21^{\circ}\ldots(ii)$
Adding $(i)$ and $(ii),$ we get
$\angle A-\angle C=63^{\circ} \ldots(iii)$
$\angle B=\angle A-42^{\circ} \ldots \ldots .[$ using $(i)]$
$\angle C=\angle A-63^{\circ} [$Using $(iii)]$
$\therefore \angle A+\angle B+\angle C=180^{\circ} [$ Sum of the angles of a triangle$] $
$\Rightarrow \angle A+\angle A-42^{\circ}+\angle A-63^{\circ}=180^{\circ}$
$\Rightarrow 3 \angle A-105^{\circ}=180^{\circ}$
$\Rightarrow 3 \angle A=285^{\circ}$
$\therefore \angle B=(95-42)^{\circ}$
$\Rightarrow \angle B=53^{\circ}$
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