PSLE CDEF and DFGH are rhombuses. EJF is a straight line.
- Find ∠FDJ.
- Find ∠EFG.
(a)
∠DFE
= (180° - 96°) ÷2
= 84° ÷ 2
= 42° (Isosceles triangle)
∠FDJ
= 180° - 121° - 42°
= 17° (Angles sum of triangle)
(b)
∠DHG
= 360° - 287°
= 73° (Angles at a point)
∠DFG = ∠DHG = 73° (Rhombus)
∠EFG
= 73° - 42°
= 31°
Answer(s): (a) 17°; (b) 31°