AC-tangent, AB-chord of a circle centered at point O, angle AOB = 146 degrees. Find the BAC angle.

Triangle AOB is isosceles, since OA and OB are circle radii.

Then the angle BAO = BAO = (180 – AOB) / 2 = (180 – 146) / 2 = 34/2 = 17.

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 – 17 = 63.

Answer: The BAC angle is 63.

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