In the figure, JKLM is a trapezium and JMNR is a rhombus. JM is parallel to KL and MN = MP. RML and MQP are straight lines. ∠MJR = 48° and ∠QMR = 29°. Find
- ∠JML
- ∠MPN
(a)
∠JMR
= (180° - 48°) ÷ 2
= 66° (Isosceles triangle)
∠JML
= 180° - 66°
= 114° (Angles on a straight line)
(b)
∠PMN
= 66° - 29°
= 37°
∠MPN
= (180° - 37°) ÷ 2
= 71.5 ° (Isosceles triangle)
Answer(s): (a) 114°; (b) 71.5°