# 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