SSC CHSL 20 Dec 2015 Shift 1 Paper

Given $\angle \mathit{B}=\frac{3}{4}\mathrm{of}\angle \mathit{A},$ ......(1)
An exterior angle of a triangle is equal to sum of the opposite interior angles.
∴ $\angle \mathit{A}+\angle \mathit{B}=112°$
Substitute equation (1) in the above equation
$\frac{4}{3}\left(\angle \mathit{B}\right)+\angle \mathit{B}=112°$
$\left(\angle \mathit{B}\right)\left(\frac{7}{3}\right)=112°$
$\angle \mathit{B}=48°$
Hence, option C is the correct answer.