PSLE In the figure, DEJK is a parallelogram and FGHJ is a rhombus. KJH is a straight line. ∠EDK = 61°, ∠JEF = 28° and ∠GFH = 32°.
- Find ∠EJF.
- Find ∠EFG.
(a)
∠EJK
= ∠EDK
= 61° (Parallelogram)
∠GHF
= ∠GFH
= 32° (Isosceles triangle)
∠FGH
= 180° - 32° - 32°
= 103° (Angles sum of triangle)
∠FJH
= ∠FGH
= 103° (Rhombus)
∠EJF
= 180° - 61° - 103°
= 16° (Angles on a straight line)
(b)
∠EFJ
= 180° - 28° - 16°
= 136° (Angles sum of triangle)
∠HFJ
= ∠GFH
= 32° (Rhombus)
∠EFG
= 360° - 136° - 32° - 32°
= 160° (Angles at a point)
Answer(s): (a) 16°; (b) 160°