PSLE In the figure, DEFG is a parallelogram. BGD and CGF are straight lines and BC = CD. ∠BCG = 24° and ∠EFG = 55°.
- Find ∠FGD.
- Find ∠GCD.
(a)
∠FGD
= 180° - 55°
= 125° (Interior angles)
(b)
∠BGC
= ∠FGD
= 125° (Vertically opposite angles)
∠CBG
= 180° - 125° - 24°
= 31° (Angles sum of triangle)
∠CDG = ∠CBG (Isosceles triangle)
∠BCD
= 180° - 31° - 31°
= 118° (Angles sum of triangle)
∠GCD
= 118° - 24°
= 94°
Answer(s): (a) 125°; (b) 94°