The figure is not drawn to scale. BCEF is a parallelogram. ∠FBO is 60° and ∠ECD is 30°. 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° - 30°
= 30° (Angles sum of triangle)
(b)
∠CBO = 30° (Isosceles triangle BCD)
∠BFE
= 180° - 60° - 30°
= 90° (Interior angles, BC//FE)
Answer(s): (a) 30°; (b) 90°