In the figure, EFGH is a quadrilateral where ∠EFG = 125° and ∠FGH = 110°. ∠GHJ = 75° and FG = GH. The point J on EH is such that FJ is parallel to GH. Calculate
- ∠FHJ
- ∠FEJ
(a)
∠GHF
= (180° - 110°) ÷ 2
= 35° (Isosceles triangle)
∠FHJ
= 75° - 35°
= 40°
(b)
∠FEJ
= 360° - 110° - 75° - 125°
= 50° (Sum of angles in a quadrilateral)
Answer(s): (a) 40°; (b) 50°