PSLE In the figure, CDHJ is a parallelogram and EFGH is a rhombus. JHG is a straight line. ∠DCJ = 60°, ∠HDE = 28° and ∠FEG = 32°.
- Find ∠DHE.
- Find ∠DEF.
(a)
∠DHJ
= ∠DCJ
= 60° (Parallelogram)
∠FGE
= ∠FEG
= 32° (Isosceles triangle)
∠EFG
= 180° - 32° - 32°
= 103° (Angles sum of triangle)
∠EHG
= ∠EFG
= 103° (Rhombus)
∠DHE
= 180° - 60° - 103°
= 17° (Angles on a straight line)
(b)
∠DEH
= 180° - 28° - 17°
= 135° (Angles sum of triangle)
∠GEH
= ∠FEG
= 32° (Rhombus)
∠DEF
= 360° - 135° - 32° - 32°
= 161° (Angles at a point)
Answer(s): (a) 17°; (b) 161°