PSLE BCDE is a rhombus. EGC and EFDH are straight lines.
- Find ∠BCE.
- Find ∠EDH.
- Find ∠EGF.
(a)
∠BCE
= (180° - 106°) ÷ 2
= 37° (Isosceles triangle)
(b)
∠CDE
= ∠CBE
= 106° (Rhombus)
∠EDH
= 180° - 106°
= 74° (Angles on a straight line)
(c)
∠FCG
= ∠BCG
= 37° (Rhombus)
∠CFG
= 180° - 69°
= 111° (Angles on a straight line)
∠EGF
= 37° + 111°
= 148° (Exterior angle of a triangle)
Answer(s): (a) 37° (b) 74° (c) 148°