PSLE In the figure, FGHJ is a parallelogram. DJF and EJH are straight lines and DE = EF. ∠DEJ = 31° and ∠GHJ = 58°.
- Find ∠HJF.
- Find ∠JEF.
(a)
∠HJF
= 180° - 58°
= 122° (Interior angles)
(b)
∠DJE
= ∠HJF
= 122° (Vertically opposite angles)
∠EDJ
= 180° - 122° - 31°
= 27° (Angles sum of triangle)
∠EFJ = ∠EDJ (Isosceles triangle)
∠DEF
= 180° - 27° - 27°
= 126° (Angles sum of triangle)
∠JEF
= 126° - 31°
= 95°
Answer(s): (a) 122°; (b) 95°