2014-II-41

Ans: C
Since $I$ is the in-centre of $\Delta QRS$, then we have

$\begin{array}{rcl}
\angle IRS & = & \angle IRQ \\
\angle IRS & = & 12^\circ
\end{array}$

Since $RS$ is the tangent to the circle at $S$, then we have

$\begin{array}{rcl}
\angle RSQ & = & \angle QPS
\end{array}$

Consider $\Delta PRS$,

$\begin{array}{rcl}
\angle RPS + \angle PSR + \angle PRS & = & 180^\circ \\
\angle QPS + 70^\circ + \angle QPS + 24^\circ & = & 180^ \circ \\
2\angle QPS & = & 86^\circ \\
\angle QPS & = & 43^\circ
\end{array}$