The figure is not drawn to scale. BCEF is a parallelogram. ∠FBO is 60° and ∠ECD is 28°. BCD is an isosceles triangle and BC = CD.
- Find ∠ODC.
- Find ∠BFE.
(a)
∠COB = 60° (Alternate angles, BF//CE)
∠COD
= 180° - 60°
= 120° (Angles on a straight line)
∠ODC
= 180° - 120° - 28°
= 32° (Angles sum of triangle)
(b)
∠CBO = 32° (Isosceles triangle BCD)
∠BFE
= 180° - 60° - 32°
= 88° (Interior angles, BC//FE)
Answer(s): (a) 32°; (b) 88°