(a) In the diagram, PQRST is a quadrilateral. PT // QS, < PTQ = 42°, < TSQ = 38° and < QSR = 30°. If < QTS = x and < POT = y, find: (i) x ; (ii) y. (b) In t...

(a) In the quadrilateral \(PT \parallel QS\), with \(\angle PTQ = 42^{\circ}\), \(\angle TSQ = 38^{\circ}\) and \(\angle QSR = 30^{\circ}\).

(i) Find \(x = \angle QTS\).

Since \(PT \parallel QS\), \(\angle PTS = \angle TSQ = 38^{\circ}\) (alternate angles). The angles at \(T\) on the straight line give:

\[ 42^{\circ} + x + 38^{\circ} = 180^{\circ}. \]

\[ 80^{\circ} + x = 180^{\circ} \quad\Rightarrow\quad x = 100^{\circ}. \]

(ii) Find \(y = \angle POT\).

Since \(PT \parallel QS\), \(\angle SQT = \angle PTQ = 42^{\circ}\) (alternate angles). In triangle \(QRS\), \(\angle SQR = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}\) (sum of angles in a triangle). The angles on the straight line at \(Q\) give:

\[ y + 42^{\circ} + 60^{\circ} = 180^{\circ}. \]

\[ y + 102^{\circ} = 180^{\circ} \quad\Rightarrow\quad y = 78^{\circ}. \]

(b) \(PQRS\) is a circle centre \(O\), with \(\angle POQ = 150^{\circ}\), \(\angle QSR = 40^{\circ}\) and \(\angle SQP = 45^{\circ}\). Find \(\angle RQS\).

The angle at the centre is twice the angle at the circumference on the same arc \(PQ\):

\[ \angle QSP = \tfrac{1}{2}\times 150^{\circ} = 75^{\circ}. \]

In triangle \(SPQ\), the sum of the angles is \(180^{\circ}\):

\[ \angle QPS + 75^{\circ} + 45^{\circ} = 180^{\circ} \quad\Rightarrow\quad \angle QPS = 60^{\circ}. \]

\(PQRS\) is a cyclic quadrilateral, so opposite angles are supplementary:

\[ \angle QRS = 180^{\circ} - \angle QPS = 180^{\circ} - 60^{\circ} = 120^{\circ}. \]

In triangle \(QRS\), the sum of the angles is \(180^{\circ}\):

\[ \angle RQS = 180^{\circ} - (\angle QRS + \angle QSR) = 180^{\circ} - (120^{\circ} + 40^{\circ}) = 20^{\circ}. \]

\(\angle RQS = 20^{\circ}.\)