In the figure, DEFG is a quadrilateral where ∠DEF = 136° and ∠EFG = 110°. ∠FGH = 71° and EF = FG. The point H on DG is such that EH is parallel to FG. Calculate
- ∠EGH
- ∠EDH
(a)
∠FGE
= (180° - 110°) ÷ 2
= 35° (Isosceles triangle)
∠EGH
= 71° - 35°
= 36°
(b)
∠EDH
= 360° - 110° - 71° - 136°
= 43° (Sum of angles in a quadrilateral)
Answer(s): (a) 36°; (b) 43°