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 = 50° and ∠LHM = 25°. Find
- ∠EHG
- ∠HKJ
(a)
∠EHM
= (180° - 50°) ÷ 2
= 65° (Isosceles triangle)
∠EHG
= 180° - 65°
= 115° (Angles on a straight line)
(b)
∠KHJ
= 65° - 25°
= 40°
∠HKJ
= (180° - 40°) ÷ 2
= 70 ° (Isosceles triangle)
Answer(s): (a) 115°; (b) 70°