PSLE CDEF is a rhombus. FHD and FGEJ are straight lines.
- Find ∠CDF.
- Find ∠FEJ.
- Find ∠FHG.
(a)
∠CDF
= (180° - 110°) ÷ 2
= 35° (Isosceles triangle)
(b)
∠DEF
= ∠DCF
= 110° (Rhombus)
∠FEJ
= 180° - 110°
= 70° (Angles on a straight line)
(c)
∠GDH
= ∠CDH
= 35° (Rhombus)
∠DGH
= 180° - 70°
= 110° (Angles on a straight line)
∠FHG
= 35° + 110°
= 145° (Exterior angle of a triangle)
Answer(s): (a) 35° (b) 70° (c) 145°