PSLE In the figure, EFGH is a parallelogram. CHE and DHG are straight lines and CD = DE. ∠CDH = 28° and ∠FGH = 57°.
- Find ∠GHE.
- Find ∠HDE.
(a)
∠GHE
= 180° - 57°
= 123° (Interior angles)
(b)
∠CHD
= ∠GHE
= 123° (Vertically opposite angles)
∠DCH
= 180° - 123° - 28°
= 29° (Angles sum of triangle)
∠DEH = ∠DCH (Isosceles triangle)
∠CDE
= 180° - 29° - 29°
= 122° (Angles sum of triangle)
∠HDE
= 122° - 28°
= 94°
Answer(s): (a) 123°; (b) 94°