The figure, not drawn to scale, O is the centre of the circle and CH // DG // EF. Find
- ∠BOE
- ∠BGO
(a)
∠HOD
= 180° - 98°
= 82° (Interior angles, CH//DO)
∠OEG = ∠OGE = 29° (Isosceles triangle)
∠DOE
= 29° + 29°
= 58° (Exterior angle of a triangle)
∠BOE
= 82° + 58°
= 140°
(b)
∠GOE
= 180° - 29° -29°
= 122° (Isosceles triangle)
∠BOG
= 360° - 140° - 122°
= 98° (Angles at a point)
BO = OG = Radius
∠BGO
= (180° - 98°) ÷ 2
= 82° ÷ 2
= 41° (Isosceles triangle)
Answer(s): (a) 140°; (b) 41°