PSLE CDEF and DFGH are rhombuses. EJF is a straight line.
- Find ∠FDJ.
- Find ∠EFG.
(a)
∠DFE
= (180° - 100°) ÷2
= 80° ÷ 2
= 40° (Isosceles triangle)
∠FDJ
= 180° - 131° - 40°
= 9° (Angles sum of triangle)
(b)
∠DHG
= 360° - 301°
= 59° (Angles at a point)
∠DFG = ∠DHG = 59° (Rhombus)
∠EFG
= 59° - 40°
= 19°
Answer(s): (a) 9°; (b) 19°