PSLE In the figure, DEFG is a parallelogram. BGD and CGF are straight lines and BC = CD. ∠BCG = 28° and ∠EFG = 56°.
- Find ∠FGD.
- Find ∠GCD.
(a)
∠FGD
= 180° - 56°
= 124° (Interior angles)
(b)
∠BGC
= ∠FGD
= 124° (Vertically opposite angles)
∠CBG
= 180° - 124° - 28°
= 28° (Angles sum of triangle)
∠CDG = ∠CBG (Isosceles triangle)
∠BCD
= 180° - 28° - 28°
= 124° (Angles sum of triangle)
∠GCD
= 124° - 28°
= 96°
Answer(s): (a) 124°; (b) 96°