Question

In Fig, O is the center of the circle. Find the angle of `triangleABC`.

Answer 1

We know that the sum of angles at a point is `360^(@)`.
`therefore angleAOB+angleBOC+angleCOA=360^(@)`
`angleAOB=360^(@)-(angleBOC + angleCOA) = 360^(@)-(110 +120)^(@)`.
`therefore angleAOB=130^(@)`.
An angle subtended by an arc at the center of the circle is double the angle subtended by the same arc at any point on the remaining circle.
`therefore angleC=1/2angleAOB=1/2(130^(@))=65^(@)`,
`angleA=1/2angleBOC=1/2(110^(@))=155^(@)` and `angleB=1/2angleAOC=1/2(120^(@))=60^(@)`.
`therefore` The angles of the triangle ABC are `angleA=55^(@), angleB=60^(@)` and `angleC=65^(@)`.