In a ∆ABC, if ∠A − ∠B = 42° and ∠B − ∠C = 21° then ∠B = ?

(c) 53°

Let 

$\angle A-\angle B=42^{\circ} \quad \ldots(i)$ and

$\angle B-\angle C=21^{\circ} \quad \ldots($ ii $)$

Adding $(i)$ and $(i i)$, we get:

$\begin{array}{ll}\angle A-\angle C=63^{\circ} & \\ \angle B=\angle A-42^{\circ} & {[\text { Using }(i)]} \\ \angle C=\angle A-63^{\circ} & {[\text { Using }(i i i)]}\end{array}$

$\therefore \angle A+\angle B+\angle C=180^{\circ} \quad$ [Sum of the angles of a triangle]

$\Rightarrow \angle A+\angle A-42^{\circ}+\angle A-63^{\circ}=180^{\circ}$

$\Rightarrow 3 \angle A-105^{\circ}=180^{\circ}$

$\Rightarrow 3 \angle A=285^{\circ}$

$\Rightarrow \angle A=95^{\circ}$

$\therefore \angle B=(95-42)^{\circ}$

$\Rightarrow \angle B=53^{\circ}$