In the figure, MNPQ is a trapezium and MQRU is a rhombus. MQ is parallel to NP and QR = QS. UQP and QTS are straight lines. ∠QMU = 45° and ∠TQU = 27°. Find
- ∠MQP
- ∠QSR
(a)
∠MQU
= (180° - 45°) ÷ 2
= 67.5° (Isosceles triangle)
∠MQP
= 180° - 67.5°
= 112.5° (Angles on a straight line)
(b)
∠SQR
= 67.5° - 27°
= 40.5°
∠QSR
= (180° - 40.5°) ÷ 2
= 69.75 ° (Isosceles triangle)
Answer(s): (a) 112.5°; (b) 69.75°