The problem asks to find the measure of arc $STP$ and arc $PQS$ in a circle, given that $\overline{PR}$ and $\overline{QT}$ are diameters of circle $A$. The measures of the central angles $\angle TAS = 39^\circ$, $\angle SAR = 51^\circ$, and $\angle PAU = 39^\circ$ are provided.

1. Problem Description

The problem asks to find the measure of arc $STP$ and arc $PQS$ in a circle, given that $\overline{PR}$ and $\overline{QT}$ are diameters of circle $A$. The measures of the central angles $\angle TAS = 39^\circ$, $\angle SAR = 51^\circ$, and $\angle PAU = 39^\circ$ are provided.

2. Solution Steps

First, we need to find the measure of arc $STP$.

$m\angle TAU = 180^{\circ} - 39^{\circ} - 51^{\circ} = 90^{\circ}$.

Therefore, $m\angle TAP = 180 - 90$.

$m\angle TAS = 39^\circ$

$m\angle TAP = 90^\circ$

Then, $m \angle SAT + m \angle TAU + m \angle UAP = m\angle STP$

$m\angle STP = 39 + 90 =129$.

Since $m \angle PAQ = 180-(90+39) = 180-129 = 51$ degrees

Since $m\angle PAU = 39^\circ$. Because $\overline{PR}$ is a diameter, the angle $\angle PAR = 180^\circ$, so $\angle RAQ = 180 - 90 - 39 = 51^\circ$

Next, we need to find the measure of arc $PQS$.

We have $m \angle PAU=39^{\circ}$. Since $\angle UAQ=90^{\circ}$ $m \angle QAR = 180-(90+39) = 51$ degrees.

Since $m\angle SAU = 90$, we can find $\angle QOA$ by the equation: $\angle QAQ =$

The measure of $\angle QOA = (180 - 39 - 51) = = 180 -129 +4 =180$. $\angle PAR= 51$

The measure of angle $PAQ =39$.

Angle $QAR = 90-39$.

We know that $PR$ is a diameter, so $m(\angle PAR) = 180$.

Then, $m(\angle RAQ) = (180-(39 +90))$ = 51

Then, $\angle PQS= PU+UQ+QS$, that corresponds to $m\angle PQS=90+39+90 =200$ (Incorrect).

$m(PQS)$ = $m(PQ) + m(QS)$. $m \angle POQ$ $180-90$

We have that $\angle R=51$, so its arc also

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1. Now from the $\angle POS$. The whole circle=

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0. $\angle POT=90=129$

$\angle ST=51 +90.$

Angle of 90+39 plus

3.

$m\angle PQS = m (\angle PAQ) + m (\angle QAS)$ = 90

3. Final Answer

m STP = 129 °

m PQS = 219 °