PSLE BCDE is a rhombus. EGC and EFDH are straight lines.
- Find ∠BCE.
- Find ∠EDH.
- Find ∠EGF.
(a)
∠BCE
= (180° - 100°) ÷ 2
= 40° (Isosceles triangle)
(b)
∠CDE
= ∠CBE
= 100° (Rhombus)
∠EDH
= 180° - 100°
= 80° (Angles on a straight line)
(c)
∠FCG
= ∠BCG
= 40° (Rhombus)
∠CFG
= 180° - 61°
= 119° (Angles on a straight line)
∠EGF
= 40° + 119°
= 159° (Exterior angle of a triangle)
Answer(s): (a) 40° (b) 80° (c) 159°