Given: circle O AB- chord angle AOB is equal to 70 degrees find: angle BAC.

Let’s draw from the point O the radii of the circle OA and OB.

Then ОА = ОВ = R, which means that the triangle ОАВ is isosceles with the base AB, which means that the angle ОАВ = ОВА = (180 – AOB) / 2 = (180 – 70) / 2 = 110/2 = 55.

Since AC is tangent to the circle, and OA is the radius drawn to the point of tangency, then OA is perpendicular to AC, which means the angle OAC = 90.

Then the angle BAC = ОАВ + ОАС = 55 + 90 = 145.

Answer: The value of the angle BAC is equal to 145.