PSLE In the figure, CDHJ is a parallelogram and EFGH is a rhombus. JHG is a straight line. ∠DCJ = 57°, ∠HDE = 29° and ∠FEG = 32°.
- Find ∠DHE.
- Find ∠DEF.
(a)
∠DHJ
= ∠DCJ
= 57° (Parallelogram)
∠FGE
= ∠FEG
= 32° (Isosceles triangle)
∠EFG
= 180° - 32° - 32°
= 106° (Angles sum of triangle)
∠EHG
= ∠EFG
= 106° (Rhombus)
∠DHE
= 180° - 57° - 106°
= 17° (Angles on a straight line)
(b)
∠DEH
= 180° - 29° - 17°
= 134° (Angles sum of triangle)
∠GEH
= ∠FEG
= 32° (Rhombus)
∠DEF
= 360° - 134° - 32° - 32°
= 162° (Angles at a point)
Answer(s): (a) 17°; (b) 162°