PSLE In the figure, DEJK is a parallelogram and FGHJ is a rhombus. KJH is a straight line. ∠EDK = 57°, ∠JEF = 28° and ∠GFH = 32°.
- Find ∠EJF.
- Find ∠EFG.
(a)
∠EJK
= ∠EDK
= 57° (Parallelogram)
∠GHF
= ∠GFH
= 32° (Isosceles triangle)
∠FGH
= 180° - 32° - 32°
= 108° (Angles sum of triangle)
∠FJH
= ∠FGH
= 108° (Rhombus)
∠EJF
= 180° - 57° - 108°
= 15° (Angles on a straight line)
(b)
∠EFJ
= 180° - 28° - 15°
= 137° (Angles sum of triangle)
∠HFJ
= ∠GFH
= 32° (Rhombus)
∠EFG
= 360° - 137° - 32° - 32°
= 159° (Angles at a point)
Answer(s): (a) 15°; (b) 159°