In the given diagram, line segment $MP$ is a tangent to circle $NQR$ at point $N$. $\angle PNQ = 64^\circ$ and $|RQ| = |RN|$. We need to find the angle marked $t$, which is $\angle RNM$.

1. Problem Description

In the given diagram, line segment $MP$ is a tangent to circle $NQR$ at point $N$. $\angle PNQ = 64^\circ$ and $|RQ| = |RN|$. We need to find the angle marked $t$, which is $\angle RNM$.

2. Solution Steps

Since $RQ = RN$, triangle $RNQ$ is an isosceles triangle. Therefore, $\angle RNQ = \angle RQN$.

The angle between a tangent and a chord is equal to the angle in the alternate segment. So, $\angle RNQ = \angle QPN = 64^\circ$.

Then $\angle RQN = \angle RNQ = 64^\circ$.

In triangle $RNQ$, the sum of the angles is $180^\circ$.

$\angle RNQ + \angle RQN + \angle NRQ = 180^\circ$

$64^\circ + 64^\circ + \angle NRQ = 180^\circ$

$128^\circ + \angle NRQ = 180^\circ$

$\angle NRQ = 180^\circ - 128^\circ = 52^\circ$.

Also, by the alternate segment theorem, $\angle QRN = 52^\circ$, which subtends the chord $QN$. The angle between the tangent $MP$ and the chord $RN$ is $t$. We can apply the tangent-chord theorem:

$\angle MNR = \angle RQN$.

$\angle MNR = t$.

Since $RQ=RN$, $\angle RQN = \angle RNQ=64^\circ$.

So $t = \angle MNR$.

We have the line $MP$, where $\angle MNQ + \angle PNQ = 180^\circ$.

So $\angle MNQ + 64^\circ= 180^\circ$.

This suggests a straight angle but this doesn't make sense.

Using the alternate segment theorem on chord $RN$ implies that $\angle QRN$ must be equal to the angle that $RN$ makes with tangent $MN$, so $t$. But we already found $\angle QRN=52^\circ$, which is incorrect. Let's use the correct theorem.

The angle between tangent and chord is equal to the angle in the alternate segment. Let $t$ be the required angle which is $\angle MNR$.

$\angle MNR = t$.

Consider triangle $RNQ$, where $RN = RQ$. Therefore, $\angle RNQ = \angle RQN$.

Given $\angle PNQ = 64^\circ$. Using the tangent chord theorem, $\angle RNQ = \angle PNQ = 64^\circ$. So, $\angle RQN = 64^\circ$.

Sum of angles in triangle $RNQ$ is $180^\circ$.

$\angle RNQ + \angle RQN + \angle QRN = 180^\circ$

$64^\circ + 64^\circ + \angle QRN = 180^\circ$

$\angle QRN = 180^\circ - 128^\circ = 52^\circ$.

Consider quadrilateral $RNMQ$.

Using the alternate segment theorem $t$ equals angle that spans $RQ$, so $t = 64$. In triangle, $RNQ$, $\angle RNQ = \angle RQN = 64^\circ$. $\angle NRQ = 52^\circ$.

Since tangent $MP$ is outside the circle, $\angle MNR+\angle PNQ=180^\circ$ does not hold here.

$NR=RQ$ means $\angle QNR = \angle NQR$. Also, $\angle QNP=64^\circ$ and it's equal to $\angle NRQ$.

Since we have $RQ=RN$, the angles opposite to it are equal.

$\angle RNQ=\angle RQN=64^{\circ}$.

$\angle QRN=180^{\circ}-64^{\circ}-64^{\circ}=52^{\circ}$.

Angle $t$ must be $\angle MNR$. $\angle QRN=52^{\circ}$

$\angle MNR=\angle RQN$ where $MN$ is the tangent.

So $\angle MNR = 64$.

3. Final Answer

D. 64°