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

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