(a) PQRST is a circle with centre C. PCS is a straight line, RS // QT, |QR| = |RS| and < QTS = 56°. Find (i) SQT (ii) PQT.
(b) In the diagram, points B and C are on a horizontal plane and |BC| = 30cm. A and D are points vertically above B and C respectively. |DC| = 40 cm and |AB| = 26 cm. Calculate the angles of depression of : (i) B from D ; (ii) A from D ; correct to the nearest degree.
Explanation
(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°)