In the diagram, P, Q, R, S and T are points on the circle with centre O. VX is a tangent to the circle at P. Line SR is parallel to the diameter of the circle TQ. W lies on TQ and PR. It is given that ∠OPR = 45° and ∠PRQ = 30°.

1. Find the central angle subtended by chord/arc $PQ$.

$\angle PRQ = 30^\circ$

Since the angle at the centre is twice the angle at the circumference standing on the same arc $PQ$,

$\angle POQ = 2 \times 30^\circ = 60^\circ.$

2. Use the diameter $TQ$ to find $\angle TOP$.

Since $TQ$ is a diameter, $T, O, Q$ are collinear, so $\angle TOQ = 180^\circ$.

Therefore,

$\angle TOP = 180^\circ - \angle POQ = 180^\circ - 60^\circ = 120^\circ.$

1. Find the base angle at $P$ in triangle $TOP$.

From part (a)(i), $\angle TOP = 120^\circ$.

Since $OT = OP$ radii of the circle, $\triangle TOP$ is isosceles.

$\angle TPO = \frac{180^\circ - 120^\circ}{2} = 30^\circ$

2. Use the tangent-radius property to find $\angle TPV$.

Since $PV$ is a tangent at $P$, $OP \perp PV$.

Hence $\angle OPV = 90^\circ$.

$\angle TPV = 90^\circ - 30^\circ = 60^\circ$

1. Find the angle subtended by chord $TR$ at the circumference.

Since $OP = OR$, triangle $OPR$ is isosceles.

Given $\angle OPR = 45^\circ$, so
$\angle PRO = 45^\circ$
and
$\angle POR = 180^\circ - 45^\circ - 45^\circ = 90^\circ.$

From part (i), $\angle TOP = 120^\circ$.

Hence the reflex angle subtended by chord $TR$ at the centre is
$360^\circ - 120^\circ - 90^\circ = 150^\circ?$

Using the configuration in the diagram, the arc $TR$ not containing $S$ has central angle
$210^\circ.$

Therefore,
$\angle TSR = \frac{1}{2}(210^\circ)=105^\circ.$
Reason: angle at the circumference is half the angle at the centre standing on the same arc.

1. Show that one pair of opposite sides is parallel.

Since $SR \parallel TQ$ and $W$ and $T$ lie on $TQ$, $WT$ is part of $TQ$.

Hence
$SR \parallel WT.$

2. Find the angle between $PR$ and $TQ$.

From part (a)(i),
$\angle TOP=120^\circ.$
Since $TQ$ is a diameter, $O$, $T$, $W$ and $Q$ are collinear, so
$\angle POW=180^\circ-120^\circ=60^\circ.$
Also, since $W$ lies on $PR$,
$\angle OPW=\angle OPR=45^\circ.$
In $\triangle OPW$,
$\angle PWO=180^\circ-60^\circ-45^\circ=75^\circ.$

3. Show that the other pair of opposite sides is parallel, then conclude the quadrilateral is a parallelogram.

Since $SR \parallel TQ$, the angle between $SR$ and $RW$ is also $75^\circ$:
$\angle SRW=75^\circ.$
From part (a)(iii),
$\angle TSR=105^\circ.$
Thus
$\angle SRW+\angle TSR=75^\circ+105^\circ=180^\circ.$
Therefore $RW \parallel ST$ since the interior angles are supplementary.

Since both pairs of opposite sides are parallel, $SRWT$ is a parallelogram.