Let us consider a circle with center O and OA and OB are two radii such that ∠AOB = 60° .

AP and BP are two intersecting tangents at point P at point A and B respectively on the circle .

To find : Angle between tangents, i.e. ∠APB

As AP and BP are tangents to given circle,

We have,

OA ⏊ AP and OB ⏊ BP

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, ∠OAP = ∠OBP = 90°

In quadrilateral AOBP,

By angle sum property of quadrilateral, we have

∠OAP + ∠OBP + ∠APB + ∠AOB = 360°

90° + 90° + ∠APB + 130° = 360°

∠APB = 50°