In a circle, when a triangle is inscribed such that one side is a diameter, the angle opposite that side is a right angle. In your case, since \(AB\) is the diameter of the circle and \(C\) is a point on the circle, then angle \(ACB\) is \(90^\circ\).
Given that the measure of arc \(BC\) is \(62^\circ\), we can find the measure of angle \(ABC\).
The key connection is that the angle at the center (which would intercept arc \(BC\)) is twice the angle at any point on the circle that intercepts the same arc. This means: \[ \text{Angle } AOB = 62^\circ \quad (\text{angle at the center of the arc}) \] Then, angle \(ABC\) is half of this: \[ \text{Angle } ABC = \frac{1}{2} (\text{arc } BC) = \frac{1}{2} \times 62^\circ = 31^\circ \]
Thus, the measure of angle \(ABC\) is 31 degrees. From the choices given, the correct answer is indeed 31 degrees.
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