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 = 48° and ∠LHM = 32°. Find
- ∠EHG
- ∠HKJ
(a)
∠EHM
= (180° - 48°) ÷ 2
= 66° (Isosceles triangle)
∠EHG
= 180° - 66°
= 114° (Angles on a straight line)
(b)
∠KHJ
= 66° - 32°
= 34°
∠HKJ
= (180° - 34°) ÷ 2
= 73 ° (Isosceles triangle)
Answer(s): (a) 114°; (b) 73°