PSLE CDEF and DFGH are rhombuses. EJF is a straight line.
- Find ∠FDJ.
- Find ∠EFG.
(a)
∠DFE
= (180° - 104°) ÷2
= 76° ÷ 2
= 38° (Isosceles triangle)
∠FDJ
= 180° - 129° - 38°
= 13° (Angles sum of triangle)
(b)
∠DHG
= 360° - 283°
= 77° (Angles at a point)
∠DFG = ∠DHG = 77° (Rhombus)
∠EFG
= 77° - 38°
= 39°
Answer(s): (a) 13°; (b) 39°