PSLE In the figure, FGHJ is a parallelogram. DJF and EJH are straight lines and DE = EF. ∠DEJ = 32° and ∠GHJ = 60°.
- Find ∠HJF.
- Find ∠JEF.
(a)
∠HJF
= 180° - 60°
= 120° (Interior angles)
(b)
∠DJE
= ∠HJF
= 120° (Vertically opposite angles)
∠EDJ
= 180° - 120° - 32°
= 28° (Angles sum of triangle)
∠EFJ = ∠EDJ (Isosceles triangle)
∠DEF
= 180° - 28° - 28°
= 124° (Angles sum of triangle)
∠JEF
= 124° - 32°
= 92°
Answer(s): (a) 120°; (b) 92°