The problem presents a diagram with a circle and some angles. Given that $\angle PMQ = 34^\circ$ and $\angle NQM = 28^\circ$, we need to find the measures of $\angle QTN$ and $\angle MPN$.

1. Problem Description

The problem presents a diagram with a circle and some angles. Given that $\angle PMQ = 34^\circ$ and $\angle NQM = 28^\circ$, we need to find the measures of $\angle QTN$ and $\angle MPN$.

2. Solution Steps

First, we will find $\angle QTN$. $\angle QTN$ is the angle at the intersection of lines QT and TN.

Since angles subtended by the same arc at the circumference of a circle are equal, we know that $\angle PNM = \angle PQM$.

Given that $\angle PQM = \angle PMQ$, we have $\angle PNM = \angle PMQ = 34^\circ$. Similarly, $\angle PTQ = \angle PNQ = 28^\circ$. Also, $\angle PTN = \angle PMN$ and $\angle MQN = \angle MPN$.

We are given that $\angle PMQ = 34^\circ$ and $\angle NQM = 28^\circ$. Then $\angle PQN = \angle PQM + \angle NQM = 34^\circ + 28^\circ = 62^\circ$.

Since $\angle QTN$ is an exterior angle of $\triangle OTQ$, we have $\angle QTN = \angle TOQ + \angle OQT = 28^\circ + \angle OPT = 28^\circ + 2 \times 34 ^\circ = 62^\circ$.

Consider triangle $NTQ$. $\angle NQT = \angle NQM = 28^\circ$. Also, $\angle TNQ = \angle TNP = \angle TMP$.

$\angle NQT = 28^{\circ}$ and $\angle NMQ = 34^\circ$. Also note that O is outside the circle. The external angle formed at O is $28^\circ$.

$\angle NOM = 2 \angle NQM = 2 \times 28^\circ = 56^\circ$

$\angle POM = 2 \angle PMQ = 2 \times 34^\circ = 68^\circ$.

Then, $\angle POQ = \angle POM + \angle MOQ = 68^\circ + 56^\circ = 124^\circ$, but $28^\circ \ne 124^\circ$.

Quadrilateral $QTNM$ is cyclic, so the exterior angle at T is equal to the interior opposite angle at N.

$\angle QTN = 180^\circ - \angle QMN$.

We also know $\angle QNM = 180^\circ - \angle QTM$.

Since points $Q, T, N, M$ lie on a circle, $QTNM$ is a cyclic quadrilateral.

Thus, the sum of the opposite angles is $180^\circ$, so $\angle QTN + \angle QMN = 180^\circ$, and $\angle TNM + \angle TQM = 180^\circ$.

Note that $\angle QMN = \angle QMP + \angle PMN$. Also $\angle QTN$ is exterior to triangle $OTQ$, so $\angle QTN = 180^\circ - (\angle T + \angle NQT)$, which is exterior to $PMN$.

Since $\angle PNM = \angle PQM = 34^\circ$ and $\angle MQN = \angle MPN$, $\angle MPN = \angle MQN = 28^\circ$.

Now for question 38, we need to find $\angle MPN$. Since $\angle MQN = \angle MPN$, $\angle MPN = 28^\circ$.

The problem is $x \rightarrow 2x + 3 = 3$.

$2x + 3 = 3$.

$2x = 0$.

$x = 0$.

3. Final Answer

For question 37, the answer is D. $118^\circ$. It should be calculated as follows:

Let angle $TOQ = a$, $\frac{62-a}{2} = 28$.

$62 - a = 56$

$a = 6$

$\angle QTN = 180^\circ - (\angle TQM + \angle NMQ)$ = $180^\circ - (34 + 28)$.

Let me come back to this.

For question 38, the answer is D. 28°.

For question 39, the answer is C.

0.