The figure, not drawn to scale, O is the centre of the circle and CH // DG // EF. Find
- ∠BOE
- ∠BGO
(a)
∠HOD
= 180° - 100°
= 80° (Interior angles, CH//DO)
∠OEG = ∠OGE = 31° (Isosceles triangle)
∠DOE
= 31° + 31°
= 62° (Exterior angle of a triangle)
∠BOE
= 80° + 62°
= 142°
(b)
∠GOE
= 180° - 31° -31°
= 118° (Isosceles triangle)
∠BOG
= 360° - 142° - 118°
= 100° (Angles at a point)
BO = OG = Radius
∠BGO
= (180° - 100°) ÷ 2
= 80° ÷ 2
= 40° (Isosceles triangle)
Answer(s): (a) 142°; (b) 40°