The distance from point M to the center O of the circle is equal to the diameter.

The distance from point M to the center O of the circle is equal to the diameter. Two straight lines are drawn through point M, tangent to the circle at points A and B. Find the angles of the triangle AOB.

1. The segments MA and MB are equal, since they are tangent segments drawn from the same point. OA and OB are equal as the radii of the circle. ∠MAO and ∠MBO are 90 ° each, because the radii OA and OB drawn to tangencies A and B are perpendicular to tangents MA and MB. ∠OMA and ∠OMB are equal, since tangents drawn from the same point make the same angles with the line connecting this point and the center of the circle.
Thus, △ MAO and △ MBO are equal right triangles.
2. Since MO is equal to the diameter, then MO is 2 times the radius of the circle, which means:
MO = 2OA = 2OB.
B △ MAO OA – leg, MO – hypotenuse. Since the OA leg is 2 times smaller than the MO hypotenuse, it lies opposite an angle of 30 °. OA lies opposite to ∠OMA, then:
∠OMA = ∠OMB = 30 °.
3. In the BMAO quadrilateral ∠MAO = ∠MBO = 90 °, ∠BMA = ∠OMA + ∠OMB = 30 ° + 30 ° = 60 °.
By the theorem on the sum of the angles of a quadrangle:
∠MAO + ∠MBO + ∠BMA + ∠AOB = 360 °;
90 ° + 90 ° + 60 ° + ∠AOB = 360 °;
∠AOB = 360 ° – 240 °;
∠AOB = 120 °.
4. В △ AOB OA = OB, then △ AOB is isosceles, ∠BAO = ∠ABO = x are the angles at the base AB.
By the theorem on the sum of the angles of a triangle:
∠BAO + ∠AOB + ∠ABO = 180 °;
x + 120 ° + x = 180 °;
2x = 180 ° – 120 °;
2x = 60 °;
x = 60 ° / 2;
x = 30 °.
So in △ AOB ∠BAO = ∠ABO = 30 °, ∠AOB = 120 °.
Answer: ∠BAO = ∠ABO = 30 °, ∠AOB = 120 °.