PSLE In the figure, DEFG is a parallelogram. BGD and CGF are straight lines and BC = CD. ∠BCG = 30° and ∠EFG = 55°.
- Find ∠FGD.
- Find ∠GCD.
(a)
∠FGD
= 180° - 55°
= 125° (Interior angles)
(b)
∠BGC
= ∠FGD
= 125° (Vertically opposite angles)
∠CBG
= 180° - 125° - 30°
= 25° (Angles sum of triangle)
∠CDG = ∠CBG (Isosceles triangle)
∠BCD
= 180° - 25° - 25°
= 130° (Angles sum of triangle)
∠GCD
= 130° - 30°
= 100°
Answer(s): (a) 125°; (b) 100°