PSLE CDEF and DFGH are rhombuses. EJF is a straight line.
- Find ∠FDJ.
- Find ∠EFG.
(a)
∠DFE
= (180° - 102°) ÷2
= 78° ÷ 2
= 39° (Isosceles triangle)
∠FDJ
= 180° - 127° - 39°
= 14° (Angles sum of triangle)
(b)
∠DHG
= 360° - 305°
= 55° (Angles at a point)
∠DFG = ∠DHG = 55° (Rhombus)
∠EFG
= 55° - 39°
= 16°
Answer(s): (a) 14°; (b) 16°