In the figure, BCDE is a parallelogram. O is the centre of the circle. Find
- ∠BED
- ∠BCE
(a)
∠OCE = ∠CEO = 25° (Angle properties within a circle)
∠EOC
= 180° - 25° - 25°
= 130° (Angles sum of triangle)
∠OBC + ∠OCB = ∠COE (Exterior angle of a triangle)
∠OBC = ∠OCB (Isosceles triangle, OB = OC)
∠OBC
= 130° ÷ 2
= 65° (Exterior angle of a triangle)
∠BED
= 180° - 65°
= 115° (Interior angles)
(b)
∠BCE
= 180° - 65° - 25°
= 90° (Angle sum of triangles)
Answer(s): (a) 115°; (b) 90°
