In the following figure,ray $QS$ is perpendicular to line $PR$. If $a: b = 7: 11$,then find $c$. Also,find reflex $\angle TQR$.

(N/A) Given that ray $QS \perp PR$,therefore $\angle PQS = \angle SQR = 90^{\circ}$.
Since $\angle PQT + \angle TQS = \angle PQS$,we have $b + a = 90^{\circ}$.
Given $a: b = 7: 11$,let $a = 7x$ and $b = 11x$.
Substituting these into the equation: $7x + 11x = 90^{\circ} \implies 18x = 90^{\circ} \implies x = 5^{\circ}$.
Thus,$a = 7 \times 5^{\circ} = 35^{\circ}$ and $b = 11 \times 5^{\circ} = 55^{\circ}$.
Since $PR$ is a straight line,$\angle PQT + \angle TQR = 180^{\circ}$ (linear pair).
$b + \angle TQR = 180^{\circ} \implies 55^{\circ} + \angle TQR = 180^{\circ} \implies \angle TQR = 125^{\circ}$.
Also,$\angle TQR$ and $\angle PQU$ are vertically opposite angles,so $\angle PQU = 125^{\circ}$.
Since $\angle PQU = \angle PQT + \angle TQU$,and $\angle PQT = b = 55^{\circ}$,then $\angle TQU = 125^{\circ} - 55^{\circ} = 70^{\circ}$.
From the figure,$\angle RQU = c$ and $\angle PQT$ and $\angle RQU$ are vertically opposite angles? No,$PR$ and $TU$ are lines intersecting at $Q$.
Thus,$\angle PQT = \angle RQU = b = 55^{\circ}$ and $\angle PQU = \angle TQR = 125^{\circ}$.
Therefore,$c = 55^{\circ}$ (vertically opposite to $b$).
Reflex $\angle TQR = 360^{\circ} - \angle TQR = 360^{\circ} - 125^{\circ} = 235^{\circ}$.