PSLE In the figure, CDHJ is a parallelogram and EFGH is a rhombus. JHG is a straight line. ∠DCJ = 59°, ∠HDE = 26° and ∠FEG = 32°.
- Find ∠DHE.
- Find ∠DEF.
(a)
∠DHJ
= ∠DCJ
= 59° (Parallelogram)
∠FGE
= ∠FEG
= 32° (Isosceles triangle)
∠EFG
= 180° - 32° - 32°
= 103° (Angles sum of triangle)
∠EHG
= ∠EFG
= 103° (Rhombus)
∠DHE
= 180° - 59° - 103°
= 18° (Angles on a straight line)
(b)
∠DEH
= 180° - 26° - 18°
= 136° (Angles sum of triangle)
∠GEH
= ∠FEG
= 32° (Rhombus)
∠DEF
= 360° - 136° - 32° - 32°
= 160° (Angles at a point)
Answer(s): (a) 18°; (b) 160°