AC-tangent, AB-chord of the circle centered at point O, angle AOB equal to 70 ° What is the angle BAC?

Since ОА = ОВ as the radii of a circle, the ABP triangle is isosceles.

Then the angle BAO = BAO = (180 – AOB) / 2 = (180 – 70) / 2 = 110/2 = 55.

By the property of a tangent drawn to a circle, the radius drawn to the tangent from the center of the circle is perpendicular to the tangent. Then the angle CAO = 90.

Angle BAC = CAB – BAO = 90 – 55 = 35.

Answer: The BAC angle is 35.