PSLE BCDE is a rhombus. EGC and EFDH are straight lines.
- Find ∠BCE.
- Find ∠EDH.
- Find ∠EGF.
(a)
∠BCE
= (180° - 110°) ÷ 2
= 35° (Isosceles triangle)
(b)
∠CDE
= ∠CBE
= 110° (Rhombus)
∠EDH
= 180° - 110°
= 70° (Angles on a straight line)
(c)
∠FCG
= ∠BCG
= 35° (Rhombus)
∠CFG
= 180° - 63°
= 117° (Angles on a straight line)
∠EGF
= 35° + 117°
= 152° (Exterior angle of a triangle)
Answer(s): (a) 35° (b) 70° (c) 152°