In the given diagram, we are given that $∠PMQ = 34°$ and $∠NQM = 28°$. We need to find the measure of $∠QTN$.

1. Problem Description

In the given diagram, we are given that $∠PMQ = 34°$ and $∠NQM = 28°$. We need to find the measure of $∠QTN$.

2. Solution Steps

Since angles $∠PMQ$ and $∠PNQ$ subtend the same arc $PQ$, we have

$∠PNQ = ∠PMQ = 34°$.

In triangle $QN$, we have $∠NQM = 28°$ and $∠PNQ = 34°$.

Therefore, $∠QNP = 34°$.

The sum of the angles in triangle $QN$ is $180°$, so

$∠NQN + ∠PNQ + ∠QNP = 180°$.

$∠NQN + 28° + 34° = 180°$.

$∠NQN + 62° = 180°$.

However, we are not looking for $∠NQN$. We are interested in $∠PTN$.

We are given that $∠NQM = 28°$ and $∠PMQ = 34°$. In triangle $QNM$, we have $∠NQM = 28°$ and $∠QMN = ∠PMQ = 34°$. The sum of angles in triangle $QNM$ is $180°$. Therefore, $∠QNM = 180° - 28° - 34° = 180° - 62° = 118°$.

$∠PTN$ and $∠PNM$ are supplementary, which means they add up to $180°$. So, $∠PTN = 180° - ∠PNM = 180° - 34° = 146°$. $∠QTM = ∠NTM$, and angles $∠NPT$ and $∠NMT$ subtend the same arc NT, so $∠NPT = ∠NMT$.

In quadrilateral $PTMN$, $∠PTN + ∠PMN = 180°$. This gives us that $∠PMN = ∠PMQ = 34°$.

The sum of the angles in quadrilateral $PTMN$ is $360°$.

In $\triangle QTN$, we want to find $∠QTN$.

We know $∠NQM = 28°$.

We also know that $∠PTM$ and $∠MTN$ are supplementary, so $∠MTN + ∠MTQ = 180°$.

$∠PTM = 180° - ∠QTM$ is part of the straight line $QT$.

We know that $∠PMT = ∠PNT$ because they subtend the same arc PT.

Consider triangle $QTP$. $∠Q = 28°$ and $∠PMQ = 34°$, $∠PTM = 180 - ∠PMQ$. Let $∠QTN$ be $x$. Then $∠PTN + ∠MTN = 180$.

However, $∠NTM = ∠NPM$ since angles subtended by same arc. We know $∠PTN = 180 - 34 = 146.$

Since $∠NQM = 28$ and $∠PMQ = 34$, in $\triangle QMN$, $∠QNM = 180 - 28 - 34 = 118$.

In cyclic quadrilateral $PTMN$, $∠PTN + ∠PMN = 180$.

$∠PMN = ∠PMQ = 34$. So $∠PTN = 180 - 34 = 146$.

So, $∠QTN = 146°$.

3. Final Answer

B. 146°