The tangents at points A and B to the circle with center O intersect at an angle of 28 °. Find the angle ABO.

Draw a segment CO.
Consider the triangle ACO.
∠ACO = ∠ACB / 2 = 28 ° / 2 = 14 ° (according to the second tangent property).
∠CAO = 90 ° (according to the first tangent property)
By the theorem on the sum of the angles of a triangle:
180 ° = ∠AOC + ∠ACO + ∠CAO
180 ° = ∠AOC + 14 ° + 90 °
∠AOC = 76 °
Consider the triangles ACO and BCO.
OC – ​​common side
AC = BC (by the second tangent property)
OA = OB (because these are the radii)
Therefore, according to the third feature, these triangles are equal.
Then ∠AOC = ∠BOC = 76 °
Consider the triangle AOB.
OA = OB (because these are the radii)
Therefore, the triangle AOB is isosceles.
Then ∠BAO = ∠ABO (by the property of an isosceles triangle).
By the theorem on the sum of the angles of a triangle:
180 ° = ∠AOB + ∠OAB + ∠ABO
180 ° = ∠AOC + ∠BOC + 2∠ABO
180 ° = 76 ° + 76 ° + 2∠ABO
28 ° = 2∠ABO
∠ABO = 14 °
Answer: 14

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