(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 the diagram, PQRS is a circle centre O. If POQ = 150°, < QSR = 40° and < SQP = 45°, calculate < RQS.
Explanation
(a)(i) < PTS = 38° (alternate angle) 42° + x + 38° = 180° (angles on a straight line) 80° + x = 180° x = 180° - 80° = 100° (ii) < SQT = 42° (alternate angle) < SQR = 60° (sum of angles in a triangle) y + 42° + 60° = 180° (angle on a straight line) y + 102° = 180° y = 180° - 102° = 78° (b) \(< QSP = \frac{150}{2}\) (angle at the centre is twicw angle at the circumference) \(< QSP = 75°\) \(\hat{R} + \hat{P} = 180°\) (opposite angles in a cyclic quad are supplementary) \(\hat{P} + 75° + 45° = 180°\) (sum of angle in a triangle) \(\hat{P} = 180° - 120°\) \(\hat{P} = 60°\) \(\hat{R} + \hat{P} = 180°\) \(\hat{R} = 180° - 60°\) \(\hat{R} = 120°\) \(< RQS + < QRS + < QSR = 180°\) (sum of angle in a triangle) \(< RQS = 180° - (40° + 120°)\) = \(180° - 160°\) \(< RQS = 20°\)