2011 WAEC Mathematics Theory (a)&nbsp; In the diagram, PQRST is a quadrilateral. PT // QS, &lt; PTQ = 42&deg;,...

(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.

(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)

\(< RQS = 180° - (40° + 120°)\)

= \(180° - 160°\)

\(< RQS = 20°\)

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