PSLE CDEF and DFGH are rhombuses. EJF is a straight line.
- Find ∠FDJ.
- Find ∠EFG.
(a)
∠DFE
= (180° - 106°) ÷2
= 74° ÷ 2
= 37° (Isosceles triangle)
∠FDJ
= 180° - 124° - 37°
= 19° (Angles sum of triangle)
(b)
∠DHG
= 360° - 297°
= 63° (Angles at a point)
∠DFG = ∠DHG = 63° (Rhombus)
∠EFG
= 63° - 37°
= 26°
Answer(s): (a) 19°; (b) 26°