# The tangents to the circle with center O at points A and B intersect at an angle of 6 °. Find the angle ABO

Draw a segment of the OS, as shown in the figure.

The triangles ACO and BCO are rectangular (by tangent property).

That is, the angles of CAO and CBO are 90 ° each.

OC – is the bisector for the angle ACB (by the property of tangents), therefore the angles ACO and BCO are 6 ° / 2 = 3 °.

By the theorem on the sum of the angles of a triangle, for a triangle ACO we write:

180 ° = ∠OAC + ∠ACO + ∠COA

180 ° = 90 ° + 3 ° + ∠COA

∠COA = 180 ° -90 ° -3 ° = 87 °

Similarly, for the triangle BCO we get that ∠COB = 87 °

∠AOB = ∠COA + ∠COB = 87 ° + 87 ° = 174 °

Draw a segment AB and consider the triangle ABO.

By the theorem on the sum of the angles of a triangle, we write:

180 ° = ∠AOB + ∠BAO + ∠ABO

180 ° = 174 ° + ∠BAO + ∠ABO

∠BAO + ∠ABO = 6 °

ABO isosceles triangle, because OA and OB are the radii of a circle and, therefore, are equal. Therefore ∠ABO = ∠BAO (by the property of an isosceles triangle). And it turns out that ∠ABO = ∠BAO = 6 ° / 2 = 3 °

Answer: ∠ABO = 3