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 = 47° and ∠QMR = 27°. Find
- ∠JML
- ∠MPN
(a)
∠JMR
= (180° - 47°) ÷ 2
= 66.5° (Isosceles triangle)
∠JML
= 180° - 66.5°
= 113.5° (Angles on a straight line)
(b)
∠PMN
= 66.5° - 27°
= 39.5°
∠MPN
= (180° - 39.5°) ÷ 2
= 70.25 ° (Isosceles triangle)
Answer(s): (a) 113.5°; (b) 70.25°