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^. (a) Find ∠DQP. (b) Find ∠ADQ.

1. Find the angle ∠BCR and ∠QCR on the straight line RCS.

We are given that $RCS$ is a straight line, $\angle BCS = 50^\circ$, and $\angle BCQ = 90^\circ$.
Therefore, the straight line angle is:
$\angle RCS = \angle RCQ + \angle BCQ + \angle BCS = 180^\circ$
$\angle RCQ + 90^\circ + 50^\circ = 180^\circ$
$\angle RCQ = 180^\circ - 140^\circ = 40^\circ$

2. Find the angle ∠RQC in triangle QRC.

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

3. Calculate angle ∠DQP on the straight line DQC.

Since $PQR$ and $DQC$ are straight lines intersecting at $Q$, the vertically opposite angle to $\angle RQC$ is $\angle DQP$.
Alternatively, because $PQR$ is a straight line, $\angle PQC = 180^\circ - \angle RQC = 180^\circ - 45^\circ = 135^\circ$.
Since $DQC$ is a straight line, $\angle DQP = 180^\circ - \angle PQC = 180^\circ - 135^\circ = 45^\circ$ is incorrect because $\angle DQP$ and $\angle RQC$ are vertically opposite angles. Let's re-verify the geometry:
$PQR$ is a straight line, so $\angle DQP + \angle DQR = 180^\circ$. Also $DQC$ is a straight line, so $\angle DQR + \angle RQC = 180^\circ$.
Thus, $\angle DQP = \angle RQC$ by vertically opposite angles (or $\angle DQP + \angle RQC = 180^\circ$ if they are supplementary on the line? No, $PQR$ is a line and $DQC$ is a line, making $\angle DQP$ and $\angle RQC$ vertically opposite angles, so $\angle DQP = \angle RQC = 45^\circ$ is incorrect since they are adjacent on the line $PQR$ and $DQC$ if they share a vertex.
Actually, the straight lines are $DQC$ and $PQR$ which intersect at $Q$. The four angles around $Q$ are:

1. $\angle DQP$
2. $\angle PQC$
3. $\angle CQR$ (which is $\angle RQC$)
4. $\angle RQD$
   By vertically opposite angles, $\angle DQP = \angle RQC = 45^\circ$. However, if we look at the diagram, $\angle DQP$ and $\angle RQC$ are vertically opposite, but let's re-examine the lines: $D, Q, C$ are on a straight line and $P, Q, R$ are on a straight line. Thus, $\angle DQP$ and $\angle RQC$ are vertically opposite angles, hence $\angle DQP = \angle RQC = 45^\circ$? No, $\angle DQP + \angle PQC = 180^\circ$ (angles on straight line $DQC$), and $\angle PQC + \angle RQC = 180^\circ$ (angles on straight line $PQR$). This means $\angle DQP = \angle RQC = 45^\circ$ is indeed correct.

1. Find the angle ∠DQP from the properties established in part (a).

First, we find $\angle QRC$ and $\angle RQC$ to find $\angle DQP$.
Since $BCS$ is a straight line, $\angle BCQ = 90^\circ$ and $\angle BCS = 50^\circ$, we have:
$\angle QCR = 180^\circ - 90^\circ - 50^\circ = 40^\circ$ (angles on a straight line).

In $\triangle QRC$, $\angle QRC = 95^\circ$ (given) and $\angle QCR = 40^\circ$.
$\angle RQC = 180^\circ - 95^\circ - 40^\circ = 45^\circ$ (angle sum of triangle).

Since $DQC$ is a straight line, $\angle DQP$ and $\angle RQC$ are vertically opposite angles because $PQR$ and $DQC$ intersect at $Q$. Therefore:
$\angle DQP = \angle RQC = 45^\circ$.

2. Find ∠ADQ using parallel lines AD and PQR.

We are given that $AD$ is parallel to $PQR$.
Since $AD \parallel PQR$, the co-interior angles (interior angles between parallel lines) must sum to $180^\circ$:
$\angle ADQ + \angle DQP = 180^\circ$

Substituting $\angle DQP = 45^\circ$:
$\angle ADQ = 180^\circ - 45^\circ = 135^\circ$