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 = 51° and ∠TQU = 25°. Find
- ∠MQP
- ∠QSR
(a)
∠MQU
= (180° - 51°) ÷ 2
= 64.5° (Isosceles triangle)
∠MQP
= 180° - 64.5°
= 115.5° (Angles on a straight line)
(b)
∠SQR
= 64.5° - 25°
= 39.5°
∠QSR
= (180° - 39.5°) ÷ 2
= 70.25 ° (Isosceles triangle)
Answer(s): (a) 115.5°; (b) 70.25°