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 = 41° and ∠HEJ = 25°. Find
- ∠BED
- ∠EGF
(a)
∠BEJ
= (180° - 41°) ÷ 2
= 69.5° (Isosceles triangle)
∠BED
= 180° - 69.5°
= 110.5° (Angles on a straight line)
(b)
∠GEF
= 69.5° - 25°
= 44.5°
∠EGF
= (180° - 44.5°) ÷ 2
= 67.75 ° (Isosceles triangle)
Answer(s): (a) 110.5°; (b) 67.75°