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 = 43° and ∠HEJ = 27°. Find
- ∠BED
- ∠EGF
(a)
∠BEJ
= (180° - 43°) ÷ 2
= 68.5° (Isosceles triangle)
∠BED
= 180° - 68.5°
= 111.5° (Angles on a straight line)
(b)
∠GEF
= 68.5° - 27°
= 41.5°
∠EGF
= (180° - 41.5°) ÷ 2
= 69.25 ° (Isosceles triangle)
Answer(s): (a) 111.5°; (b) 69.25°