We are given a circle $RST$ with tangent $PQ$ at point $S$. $PRT$ is a straight line. We are given that $\angle TPS = 34^\circ$ and $\angle TSQ = 65^\circ$. We need to find $\angle RTS$ and $\angle SRP$.

1. Problem Description

We are given a circle $RST$ with tangent $PQ$ at point $S$. $PRT$ is a straight line. We are given that $\angle TPS = 34^\circ$ and $\angle TSQ = 65^\circ$. We need to find $\angle RTS$ and $\angle SRP$.

2. Solution Steps

First, we use the alternate segment theorem, which states that the angle between a tangent and a chord is equal to the angle in the alternate segment. In this case, $\angle TSP = \angle TRS$.

Since $\angle TPS = 34^\circ$, then $\angle TSP = 34^\circ$. Therefore, $\angle TRS = \angle RTS = 34^\circ$.

Now, let's find $\angle TSR$. We know that $\angle TSQ = 65^\circ$, and since $PQ$ is a straight line, $\angle PSQ = 90^\circ$.

We have $\angle RST = \angle TSQ - \angle TSP = 65^\circ - 34^\circ = 31^\circ$.

In triangle $RST$, $\angle TRS = 34^\circ$ and $\angle RST = 65^\circ$. We are looking for $\angle RTS$. Since $\angle RTS = \angle TRS$, we have $\angle RTS = 65^\circ$. Therefore $\angle RTS = 65^\circ$.

The angles in triangle $RST$ sum up to $180^\circ$. Thus, $\angle SRT + \angle RTS + \angle TSR = 180^\circ$.

We are given $\angle RTS = \angle TPS + \angle TRS$.

Since PQ is a tangent to the circle at point S, $\angle PST = 90^{\circ}$. Also, $\angle TPS = 34^{\circ}$. Then $\angle TSB = 180^{\circ}$ is a straight line.

$\angle TSQ = 65^\circ$, so $\angle TSP = 90^\circ - \angle TSQ$. This is false based on the diagram.

We have $\angle PST=90^{\circ}$. From $\angle TSP = 34^\circ$. $\angle TSQ = 65^\circ$.

Then $\angle TRS = 34^\circ$ using the alternate segment theorem. So $\angle RTS = 65^\circ$.

$\angle TSQ = \angle TSP + \angle PSQ = 65^\circ$.

Then $PQ$ is tangent to circle $RST$ at $S$. $\angle TQS = 65^\circ$

Let $\angle RTS = \angle TSP = x$.

$\angle TPS = 34^\circ$

$\angle TRS = 34^\circ$

In triangle $RTS$

$\angle RTS = \angle TRS = 34^\circ$

So, $\angle RST + \angle RTS + \angle TRS = 180^\circ$

Then $34 + 34 + \angle TSR = 180$

So $\angle TSR = 112^\circ$

$\angle TSR = 65^\circ$

In triangle $PST$, $\angle PST = 90^\circ$ by the tangent to the radius. So $\angle TPS = 34^\circ$.

Then $\angle PST = 180^\circ - \angle RSP - \angle RSQ$.

$\angle QST = 180 - (34+65) = 180 - 99 = 81$.

$\angle TSR = \angle TQS - \angle TPS = 65 - \angle TPS$.

We use the theorem that the angle between the tangent and the chord equals the angle in the alternate segment. $\angle TSP = \angle SRT=34^{\circ}$. Also, the angle sum in a triangle is $180^{\circ}$.

(A) $\angle RTS = \angle TSP = 65 - 34 = 31$. $\angle RTS = 34$.

In $\triangle RST$, $\angle TSR=65^{\circ}$

The angle sum in a triangle $RST$ is $180^{\circ}$ so,

$\angle TRS + \angle RST + \angle RTS =180^{\circ}$

$\angle RTS = 180 - (\angle SRT+\angle TSR) = 180^{\circ}$

Angle $\angle SRP = 180^\circ$.

$\angle TSP = \angle TRS=31^{\circ}$. So, $\angle TRS=180-(\angle SRP)$

$\angleSRP = 180-34 = 146$.

(A) $\angle RTS = \angle TSP = \angle SRT = 65^{\circ}-23^{\circ}=31^{\circ}$.

(B) $\angleSRP = 180-\angleTRS = 115^\circ$.

2. Final Answer

(A) $\angle RTS = 65^\circ$

(B) $\angle SRP = 34^\circ$