PSLE CDEF is a rhombus. FHD and FGEJ are straight lines.
- Find ∠CDF.
- Find ∠FEJ.
- Find ∠FHG.
(a)
∠CDF
= (180° - 100°) ÷ 2
= 40° (Isosceles triangle)
(b)
∠DEF
= ∠DCF
= 100° (Rhombus)
∠FEJ
= 180° - 100°
= 80° (Angles on a straight line)
(c)
∠GDH
= ∠CDH
= 40° (Rhombus)
∠DGH
= 180° - 62°
= 118° (Angles on a straight line)
∠FHG
= 40° + 118°
= 158° (Exterior angle of a triangle)
Answer(s): (a) 40° (b) 80° (c) 158°