# 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 ° Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.