(a) From the figure < QRS = 180^\circ - 56^\circ = 124^\circ (opposite angles of a parallelogram) \(\therefore\) < RQS = < PQS = \(\frac{180^\circ - 124^\circ}{2} = 28^\circ\) (base angles of an isosceles triangle) \(\therefore\) < SQT = < RSQ = 28^\circ (alternate angles PS // QT) (ii) < PQT = 90^\circ - 28^\circ = 62^\circ (< PQS is in a semi-circle) (b) \(\tan < BDC = \frac{30}{40} = 0.75\) \(< BDC = \tan^{-1} (0.75) = 36.87^\circ\) \(\therefore < BDF = 90^\circ - 36.87^\circ = 53.13^\circ \approxeq 53^\circ\) \(\tan < ADF = \frac{14}{30} = 0.4667\) \(< ADF = \tan^{-1} (0.4667) = 25.02^\circ \approxeq 25^\circ\)