PSLE CDEF and DFGH are rhombuses. EJF is a straight line.
- Find ∠FDJ.
- Find ∠EFG.
(a)
∠DFE
= (180° - 108°) ÷2
= 72° ÷ 2
= 36° (Isosceles triangle)
∠FDJ
= 180° - 126° - 36°
= 18° (Angles sum of triangle)
(b)
∠DHG
= 360° - 278°
= 82° (Angles at a point)
∠DFG = ∠DHG = 82° (Rhombus)
∠EFG
= 82° - 36°
= 46°
Answer(s): (a) 18°; (b) 46°