PQRS are points on the circumference of a circle, centre O. Angle OQR = 50°, angle PSR = 73° and obtuse angle SOQ = 156°.

1. Find the angle at the circumference subtended by chord/arc $SQ$.

$\angle QRS = \dfrac{1}{2}\angle SOQ$
$= \dfrac{1}{2}(156^\circ)$
$= 78^\circ$

Reason: angle at the centre is twice the angle at the circumference standing on the same arc.

1. Find the central angle $POQ$.

Since $OQ = OR$, triangle $OQR$ is isosceles.

$\angle OQR = \angle ORQ = 50^\circ$

So
$\angle QOR = 180^\circ - 50^\circ - 50^\circ = 80^\circ.$

Also, $\angle PSR = 73^\circ$, so the angle at the centre standing on the same arc $PR$ is
$\angle POR = 2(73^\circ)=146^\circ.$

Hence
$\angle POQ = \angle POR - \angle QOR = 146^\circ - 80^\circ = 66^\circ.$

2. Find $\angle PSO$ using triangle $POS$.

Given the obtuse angle
$\angle SOQ = 156^\circ.$

Since $\angle SOQ = \angle SOP + \angle POQ$,
$\angle SOP = 156^\circ - 66^\circ = 90^\circ.$

In triangle $POS$, $OP = OS$, so it is isosceles.

Therefore
$\angle PSO = \frac{180^\circ - 90^\circ}{2} = 45^\circ.$

1. Use the equal radii to identify triangle $POQ$ as isosceles.

From part (ii), $\angle POQ = 66^\circ$.
Since $OP = OQ$ because they are radii of the same circle, triangle $POQ$ is isosceles.
Hence $\angle OPQ = \angle PQO$.

2. Find each base angle of triangle $POQ$.

$\angle PQO = \frac{180^\circ - 66^\circ}{2} = 57^\circ$

1. Find the interior angle $\angle PQR$.

From part (iii), $\angle PQO = 57^\circ$.

Given $\angle OQR = 50^\circ$,
$\angle PQR = \angle PQO + \angle OQR = 57^\circ + 50^\circ = 107^\circ.$

2. Use co-interior angles to decide whether $PQ$ is parallel to $SR$.

From part (i), $\angle QRS = 78^\circ$.

If $PQ \parallel SR$, then the co-interior angles on transversal $QR$ should add to $180^\circ$.

But
$\angle PQR + \angle QRS = 107^\circ + 78^\circ = 185^\circ \ne 180^\circ.$

Therefore, $PQ$ is not parallel to $SR$.