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 = 45° and ∠LHM = 32°. Find
- ∠EHG
- ∠HKJ
(a)
∠EHM
= (180° - 45°) ÷ 2
= 67.5° (Isosceles triangle)
∠EHG
= 180° - 67.5°
= 112.5° (Angles on a straight line)
(b)
∠KHJ
= 67.5° - 32°
= 35.5°
∠HKJ
= (180° - 35.5°) ÷ 2
= 72.25 ° (Isosceles triangle)
Answer(s): (a) 112.5°; (b) 72.25°