PSLE BCDE is a rhombus. EGC and EFDH are straight lines.
- Find ∠BCE.
- Find ∠EDH.
- Find ∠EGF.
(a)
∠BCE
= (180° - 104°) ÷ 2
= 38° (Isosceles triangle)
(b)
∠CDE
= ∠CBE
= 104° (Rhombus)
∠EDH
= 180° - 104°
= 76° (Angles on a straight line)
(c)
∠FCG
= ∠BCG
= 38° (Rhombus)
∠CFG
= 180° - 64°
= 116° (Angles on a straight line)
∠EGF
= 38° + 116°
= 154° (Exterior angle of a triangle)
Answer(s): (a) 38° (b) 76° (c) 154°