In the figure, YBEH is a rectangle, EFHJ is a rhombus and BCDE is a parallelogram. ∠EFH = 72° and ∠DEF = 92°.
- Find ∠AEB
- Find ∠EDC
(a)
∠HJE = ∠HFE = 72°
∠JEH
= (180° - 72°) ÷ 2
= 108° ÷ 2
= 54° (Isosceles triangle)
∠AEB
= 90° - 54°
= 36°
(b)
∠BED
= 360° - 92° - 90° - 54°
= 124° (Angles at a point)
∠CDE
= 180° - 124°
= 56° (Interior Angles, BE//CF)
Answer(s): (a) 36°; (b) 56°