In the figure, BCDE is a trapezium and BEFJ is a rhombus. BE is parallel to CD and EF = EG. JED and EHG are straight lines. ∠EBJ = 40° and ∠HEJ = 33°. Find
- ∠BED
- ∠EGF
(a)
∠BEJ
= (180° - 40°) ÷ 2
= 70° (Isosceles triangle)
∠BED
= 180° - 70°
= 110° (Angles on a straight line)
(b)
∠GEF
= 70° - 33°
= 37°
∠EGF
= (180° - 37°) ÷ 2
= 71.5 ° (Isosceles triangle)
Answer(s): (a) 110°; (b) 71.5°