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 = 51° and ∠HEJ = 35°. Find
- ∠BED
- ∠EGF
(a)
∠BEJ
= (180° - 51°) ÷ 2
= 64.5° (Isosceles triangle)
∠BED
= 180° - 64.5°
= 115.5° (Angles on a straight line)
(b)
∠GEF
= 64.5° - 35°
= 29.5°
∠EGF
= (180° - 29.5°) ÷ 2
= 75.25 ° (Isosceles triangle)
Answer(s): (a) 115.5°; (b) 75.25°