PSLE In the figure, EFGH is a parallelogram. CHE and DHG are straight lines and CD = DE. ∠CDH = 31° and ∠FGH = 60°.
- Find ∠GHE.
- Find ∠HDE.
(a)
∠GHE
= 180° - 60°
= 120° (Interior angles)
(b)
∠CHD
= ∠GHE
= 120° (Vertically opposite angles)
∠DCH
= 180° - 120° - 31°
= 29° (Angles sum of triangle)
∠DEH = ∠DCH (Isosceles triangle)
∠CDE
= 180° - 29° - 29°
= 122° (Angles sum of triangle)
∠HDE
= 122° - 31°
= 91°
Answer(s): (a) 120°; (b) 91°