The diagram shows a circle with center $O$. $MN$ is a tangent to the circle at point $S$. The angle between the tangent $MN$ and the chord $PS$ is $55^\circ$. We need to find the measure of angle $\angle RPS$.

1. Problem Description

The diagram shows a circle with center $O$. $MN$ is a tangent to the circle at point $S$. The angle between the tangent $MN$ and the chord $PS$ is $55^\circ$. We need to find the measure of angle $\angle RPS$.

2. Solution Steps

We know that the angle between a tangent and a chord is equal to the angle subtended by the chord in the alternate segment. Therefore, $\angle RSP = 55^\circ$.

Since $O$ is the center of the circle, $OS$ is perpendicular to the tangent $MN$ at the point of tangency $S$. So, $\angle OSN = 90^\circ$.

Consider the triangle $OSP$. Since $OS$ and $OP$ are both radii of the circle, $OS = OP$. Therefore, triangle $OSP$ is an isosceles triangle. Thus, $\angle OPS = \angle OSP$.

Since $\angle PSN = 55^\circ$ and $\angle OSN = 90^\circ$, $\angle OSP = 90^\circ - 55^\circ = 35^\circ$.

Since $\angle OPS = \angle OSP$, $\angle OPS = 35^\circ$.

The angle subtended by the arc $PR$ at the center is $\angle POS$. Since the sum of angles in triangle $OSP$ is $180^\circ$, we have $\angle OPS + \angle OSP + \angle POS = 180^\circ$.

Therefore, $35^\circ + 35^\circ + \angle POS = 180^\circ$, which gives us $\angle POS = 180^\circ - 70^\circ = 110^\circ$.

The angle at the circumference, $\angle PRS$, is half the angle at the center, $\angle POS$. Therefore, $\angle PRS = \frac{1}{2} \angle POS = \frac{1}{2} \times 110^\circ = 55^\circ$.

We know $\angle RSP = 55^\circ$ and $\angle OPS = 35^\circ$.

In triangle $PSR$, the sum of the angles is $180^\circ$. $\angle RPS + \angle RSP + \angle PSR = 180^\circ$.

We are seeking to find $\angle RPS$.

Let $\angle RPS = x$.

We know that $\angle RSP = 55^\circ$. Since $PS$ and $SR$ do not necessarily have to be equal, the angles would change.

The theorem we need is the angle subtended by a chord at the tangent is equal to the angle in the alternate segment. That is $\angle PSM = \angle PRS = 55^\circ$.

Now consider triangle $RPS$. $\angle RPS + \angle PSR + \angle SRP = 180^\circ$.

So, $\angle RPS + 55^\circ + \angle SRP = 180^\circ$.

The angle $\angle RPS$ is subtended by the arc $RS$. The angle at the circumference $\angle PRS = 55^\circ$. Since $OS \perp MN$, we have $\angle OSP = 90^\circ - 55^\circ = 35^\circ$. Also $OP=OS$, so $\angle OPS = 35^\circ$.

Now consider triangle $OPS$. Then $\angle POS = 180^\circ - (35^\circ+35^\circ) = 180^\circ - 70^\circ = 110^\circ$.

The angle subtended by the arc $PS$ at the remaining part of the circle at $R$ is $\angle PRS = \frac{1}{2} \angle POS = \frac{110^\circ}{2} = 55^\circ$.

Finally, we know that $\angle RPS = |35^\circ - 55^\circ| = |-20^\circ|$.

$\angle PRS = 35^\circ$

3. Final Answer

The final answer is $\boxed{35°}$