The angle between chord AB and tangent BC is 32 degrees. Find the value of the smaller arc contracted by the chord AB.

Since BC is tangent to the circle, the OB radius drawn to the tangency point is perpendicular to this tangent. Then the OBC angle is 90.

Angle ОВА = (ОВС – ВС) = (90 – 32) = 48.

The AOB triangle is isosceles, since its two sides are circular radii, then the angle OBA = OAB = 480, and the angle AOB = (180 – 48 – 48) = 104.

Since the angle AOB is central and rests on the arc AB, the degree mega of the smaller arc AB is equal to the central angle AOB.

The arc AB is equal to 104.

Answer: The magnitude of the smaller arc is 104.