PSLE In the figure, EFGH is a parallelogram. CHE and DHG are straight lines and CD = DE. ∠CDH = 31° and ∠FGH = 58°.
- Find ∠GHE.
- Find ∠HDE.
(a)
∠GHE
= 180° - 58°
= 122° (Interior angles)
(b)
∠CHD
= ∠GHE
= 122° (Vertically opposite angles)
∠DCH
= 180° - 122° - 31°
= 27° (Angles sum of triangle)
∠DEH = ∠DCH (Isosceles triangle)
∠CDE
= 180° - 27° - 27°
= 126° (Angles sum of triangle)
∠HDE
= 126° - 31°
= 95°
Answer(s): (a) 122°; (b) 95°