PSLE In the figure, EFGH is a parallelogram. CHE and DHG are straight lines and CD = DE. ∠CDH = 31° and ∠FGH = 59°.
- Find ∠GHE.
- Find ∠HDE.
(a)
∠GHE
= 180° - 59°
= 121° (Interior angles)
(b)
∠CHD
= ∠GHE
= 121° (Vertically opposite angles)
∠DCH
= 180° - 121° - 31°
= 28° (Angles sum of triangle)
∠DEH = ∠DCH (Isosceles triangle)
∠CDE
= 180° - 28° - 28°
= 124° (Angles sum of triangle)
∠HDE
= 124° - 31°
= 93°
Answer(s): (a) 121°; (b) 93°