The figure, not drawn to scale, O is the centre of the circle and CH // DG // EF. Find
- ∠BOE
- ∠BGO
(a)
∠HOD
= 180° - 94°
= 86° (Interior angles, CH//DO)
∠OEG = ∠OGE = 33° (Isosceles triangle)
∠DOE
= 33° + 33°
= 66° (Exterior angle of a triangle)
∠BOE
= 86° + 66°
= 152°
(b)
∠GOE
= 180° - 33° -33°
= 114° (Isosceles triangle)
∠BOG
= 360° - 152° - 114°
= 94° (Angles at a point)
BO = OG = Radius
∠BGO
= (180° - 94°) ÷ 2
= 86° ÷ 2
= 43° (Isosceles triangle)
Answer(s): (a) 152°; (b) 43°