If a chord `AB` subtends and angle of `60^(@)` at the centre of a circle, then the angle between the tangents to the circle drawn from `A` and `B` is

Explanation

Since a chord AB subtends an angle of `60^(@)` at the centre of a circle.
i.e., `angleAOB=60^(@)`
As OA = OB= Radius of the circle
`angleAOB=60^(@)`
The tangent at pionts A and B is drawn, which intersect at C.
We know, `OAbotACand OBbotBC`.
`:.angleOAC=90^(@),angleOBC=90^(@)`
`rArrangleOAB+angleBAC=90^(@)`
and `rArrangleOBA+angleABC=90^(@)` `rArrangleBAC=90^(@)-60^(@)=30^(@)`
and `rArrangleABC=90^(@)-60^(@)=30^(@)`
In `DeltaABC`, `angleBAC+angleCBA+angleACB=180^(@)`
[since, sum of all interior angles of a triangle is `180^(@)`]
`rArrangleACB=180^(@)-(30^(@)+30^(@))=120^(@)`