PSLE In the figure, CDHJ is a parallelogram and EFGH is a rhombus. JHG is a straight line. ∠DCJ = 61°, ∠HDE = 29° and ∠FEG = 32°.
- Find ∠DHE.
- Find ∠DEF.
(a)
∠DHJ
= ∠DCJ
= 61° (Parallelogram)
∠FGE
= ∠FEG
= 32° (Isosceles triangle)
∠EFG
= 180° - 32° - 32°
= 103° (Angles sum of triangle)
∠EHG
= ∠EFG
= 103° (Rhombus)
∠DHE
= 180° - 61° - 103°
= 16° (Angles on a straight line)
(b)
∠DEH
= 180° - 29° - 16°
= 135° (Angles sum of triangle)
∠GEH
= ∠FEG
= 32° (Rhombus)
∠DEF
= 360° - 135° - 32° - 32°
= 161° (Angles at a point)
Answer(s): (a) 16°; (b) 161°