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