We are given a circle with center O. We are given the measure of the inscribed angle $\angle PRQ = 48^{\circ}$. We need to find the measure of the angle $\angle QPR$.

1. Problem Description

We are given a circle with center O. We are given the measure of the inscribed angle $\angle PRQ = 48^{\circ}$. We need to find the measure of the angle $\angle QPR$.

2. Solution Steps

The measure of a central angle is twice the measure of an inscribed angle that intercepts the same arc. The central angle corresponding to inscribed angle $\angle PRQ$ is $\angle POQ$. So we have:

$m\angle POQ = 2 \cdot m\angle PRQ$

$m\angle POQ = 2 \cdot 48^{\circ} = 96^{\circ}$.

Since $OQ$ and $OP$ are both radii of the circle, triangle $OPQ$ is an isosceles triangle with $OQ = OP$. Therefore, the angles opposite to these equal sides must be equal: $\angle OQP = \angle OPQ$.

The sum of the angles in a triangle is $180^{\circ}$, so in triangle $OPQ$, we have

$m\angle POQ + m\angle OPQ + m\angle OQP = 180^{\circ}$.

$96^{\circ} + m\angle OPQ + m\angle OQP = 180^{\circ}$.

Since $m\angle OPQ = m\angle OQP$, we can rewrite this as:

$96^{\circ} + 2 \cdot m\angle OPQ = 180^{\circ}$.

$2 \cdot m\angle OPQ = 180^{\circ} - 96^{\circ}$

$2 \cdot m\angle OPQ = 84^{\circ}$

$m\angle OPQ = \frac{84^{\circ}}{2}$

$m\angle OPQ = 42^{\circ}$.

Therefore, $m\angle QPR = 42^{\circ}$.

3. Final Answer

$42$