In the figure below, ABCD is a trapezium. APB, DQC, RCS and PQR are straight lines. APB is parallel to DQC and AD is parallel to PQR. ∠BCS = 50°, ∠BCQ = 90° and ∠QRC = 95°.

1. Find $\angle QCS$ and then $\angle QCR$

$\angle QCS = \angle BCQ + \angle BCS = 90^\circ + 50^\circ = 140^\circ$
Since $RCS$ is a straight line, angles on a straight line add up to $180^\circ$:
$\angle QCR = 180^\circ - 140^\circ = 40^\circ$

2. Find $\angle RQC$ using the angle sum of $\triangle RQC$

In $\triangle RQC$, the sum of angles is $180^\circ$:
$\angle RQC = 180^\circ - \angle QRC - \angle QCR$
$\angle RQC = 180^\circ - 95^\circ - 40^\circ = 45^\circ$

3. Find $\angle DQP$ using vertically opposite angles

Since $DQC$ and $PQR$ are straight lines, $\angle DQP$ and $\angle RQC$ are vertically opposite angles.
$\angle DQP = \angle RQC = 45^\circ$

1. Find $\angle QCR$ using the straight line RCS

Since $R, C, S$ are collinear, the angles on the straight line at $C$ sum to $180^\circ$.
$\angle QCR = 180^\circ - \angle BCQ - \angle BCS$
$\angle QCR = 180^\circ - 90^\circ - 50^\circ = 40^\circ$

2. Find $\angle RQC$ using the angle sum of $\triangle QRC$

In $\triangle QRC$:
$\angle RQC = 180^\circ - \angle QRC - \angle QCR$
$\angle RQC = 180^\circ - 95^\circ - 40^\circ = 45^\circ$

3. Find $\angle DQP$ (the answer to part a) using vertically opposite angles

Since $D, Q, C$ and $P, Q, R$ are straight lines:
$\angle DQP = \angle RQC = 45^\circ$ (vertically opposite angles)

4. Find $\angle ADQ$ using alternate angles

Given that $AD \parallel PQR$, and $DQC$ is a transversal:
$\angle ADQ = \angle DQP$ (alternate angles, $AD \parallel PQR$)
$\angle ADQ = 45^\circ$