In the figure, EFGH is a trapezium and EHJM is a rhombus. EH is parallel to FG and HJ = HK. MHG and HLK are straight lines. ∠HEM = 46° and ∠LHM = 31°. Find
- ∠EHG
- ∠HKJ
(a)
∠EHM
= (180° - 46°) ÷ 2
= 67° (Isosceles triangle)
∠EHG
= 180° - 67°
= 113° (Angles on a straight line)
(b)
∠KHJ
= 67° - 31°
= 36°
∠HKJ
= (180° - 36°) ÷ 2
= 72 ° (Isosceles triangle)
Answer(s): (a) 113°; (b) 72°