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 = 34° (Isosceles triangle)
∠DOE
= 34° + 34°
= 68° (Exterior angle of a triangle)
∠BOE
= 86° + 68°
= 154°
(b)
∠GOE
= 180° - 34° -34°
= 112° (Isosceles triangle)
∠BOG
= 360° - 154° - 112°
= 94° (Angles at a point)
BO = OG = Radius
∠BGO
= (180° - 94°) ÷ 2
= 86° ÷ 2
= 43° (Isosceles triangle)
Answer(s): (a) 154°; (b) 43°
