Question

2 и∂ σиє ρℓѕѕѕ.......нєℓρ мє ιи тнιѕ.....

Answer 1

i) In triangle OBC,

∠OBC = ∠OCB ( isoceles triangle)

let ∠BOC be x

∴30°+30°+x= 180° (angle sum property)

60°+x=180°

∴x= 180°-60°

x= 120°

ii) ∠BOC + ∠BOA = 180° (linear angles)

∴ 120° + ∠BOA = 180°

∠BOA = 180° - 120°

           = 60°

iii) in triangle OBA,

∠BOA = 60°

∠OBA = 90° (the angle between a tangent and the radius at the point of                                                  contact is 90°)

∴ ∠BOA + ∠OBA + ∠OAB = 180°

⇒ 90° + 60° + x = 180°

x = 180° - (90° + 60°)

x = 180° - 150°

∴x = 30°

the answer is 30°

please mark as brainliest answer if correct!