PSLE In the figure, CDHJ is a parallelogram and EFGH is a rhombus. JHG is a straight line. ∠DCJ = 56°, ∠HDE = 28° and ∠FEG = 33°.
- Find ∠DHE.
- Find ∠DEF.
(a)
∠DHJ
= ∠DCJ
= 56° (Parallelogram)
∠FGE
= ∠FEG
= 33° (Isosceles triangle)
∠EFG
= 180° - 33° - 33°
= 104° (Angles sum of triangle)
∠EHG
= ∠EFG
= 104° (Rhombus)
∠DHE
= 180° - 56° - 104°
= 20° (Angles on a straight line)
(b)
∠DEH
= 180° - 28° - 20°
= 132° (Angles sum of triangle)
∠GEH
= ∠FEG
= 33° (Rhombus)
∠DEF
= 360° - 132° - 33° - 33°
= 162° (Angles at a point)
Answer(s): (a) 20°; (b) 162°