(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. (SEE THE DIAGRAMS ABOVE)
(b)
In the diagram, PQRS is a circle centre O. If POQ = 150°, < QSR = 40° and < SQP = 45°, calculate < RQS.
Attached Figure / Diagram
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a)(i) < PTS = 38^\circ (alternate angle)
42^\circ + x + 38^\circ = 180^\circ (angles on a straight line)
80^\circ + x = 180^\circ
x = 180^\circ - 80^\circ = 100^\circ
(ii) < SQT = 42^\circ (alternate angle)
< SQR = 60^\circ (sum of angles in a triangle)
y + 42^\circ + 60^\circ = 180^\circ (angle on a straight line)
y + 102^\circ = 180^\circ
y = 180^\circ - 102^\circ = 78^\circ
(b) \(< QSP = \frac{150}{2}\) (angle at the centre is twicw angle at the circumference)
\(< QSP = 75^\circ\)
\(\hat{R} + \hat{P} = 180^\circ\) (opposite angles in a cyclic quad are supplementary)
\(\hat{P} + 75^\circ + 45^\circ = 180^\circ\) (sum of angle in a triangle)
\(\hat{P} = 180^\circ - 120^\circ\)
\(\hat{P} = 60^\circ\)
\(\hat{R} + \hat{P} = 180^\circ\)
\(\hat{R} = 180^\circ - 60^\circ\)
\(\hat{R} = 120^\circ\)
\(< RQS + < QRS + < QSR = 180^\circ\) (sum of angle in a triangle)
\(< RQS = 180^\circ - (40^\circ + 120^\circ)\)
= \(180^\circ - 160^\circ\)
\(< RQS = 20^\circ\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #52748)