PSLE BCDE is a rhombus. EGC and EFDH are straight lines.
- Find ∠BCE.
- Find ∠EDH.
- Find ∠EGF.
(a)
∠BCE
= (180° - 108°) ÷ 2
= 36° (Isosceles triangle)
(b)
∠CDE
= ∠CBE
= 108° (Rhombus)
∠EDH
= 180° - 108°
= 72° (Angles on a straight line)
(c)
∠FCG
= ∠BCG
= 36° (Rhombus)
∠CFG
= 180° - 61°
= 119° (Angles on a straight line)
∠EGF
= 36° + 119°
= 155° (Exterior angle of a triangle)
Answer(s): (a) 36° (b) 72° (c) 155°