Question

whats the size of \\( \angle abc \\) shown in the above figure?

Explanation:

Step1: Recall the central - angle and inscribed - angle relationship

The measure of an inscribed angle is half of the measure of the central angle that subtends the same arc.

Step2: Identify the central angle and the inscribed angle

The central angle \(\angle BOC\) and the inscribed angle \(\angle BAC\) (assuming \(A\) is a point on the circumference such that \(\angle ABC\) is related to the arc \(BC\)). Wait, no, actually, for the circle with center \(O\), the central angle \(\angle BOC\) and the inscribed angle \(\angle BAC\) subtend the same arc \(BC\). But if we consider the property that the angle between a tangent and a chord is equal to the inscribed angle in the alternate segment. However, if we assume the standard central - inscribed angle relation:
Let the central angle \(\angle BOC = 70^{\circ}\). The inscribed angle \(\angle BAC=\frac{1}{2}\angle BOC\). But if we consider the exterior angle property of a triangle. Wait, no, another approach:
We know that \(OB = OC\) (radii of the same circle). Let's assume the question is about the angle subtended by an arc. The measure of \(\angle ABC\) (assuming \(A\) is a point on the circumference such that \(\angle ABC\) is an inscribed angle). The central angle \(\angle BOC = 70^{\circ}\). The inscribed angle \(\angle BAC=\frac{1}{2}\angle BOC\). But if we use the property that the angle between a tangent and a chord is equal to the inscribed angle in the alternate segment. Wait, no, if we consider the basic central - inscribed angle formula:
The measure of an inscribed angle \(\theta_{i}\) and central angle \(\theta_{c}\) subtending the same arc: \(\theta_{i}=\frac{1}{2}\theta_{c}\).
If we assume that the angle \(\angle ABC\) is an inscribed angle and the central angle for the same arc \(AC\) (no, wait, looking at the figure again, if we consider the property that \(\angle ABC=\frac{1}{2}(180^{\circ}-\angle BOC)\) (using the cyclic quadrilateral property or the property of angles in a circle). Wait, no, another way:
We know that \(OB = OC\), so \(\triangle BOC\) is isosceles. \(\angle OBC=\angle OCB=\frac{180^{\circ}-\angle BOC}{2}=\frac{180 - 70}{2}=55^{\circ}\).
The angle between the tangent (if \(BA\) is a tangent) and the chord \(BC\) is equal to the inscribed angle in the alternate segment. But if we assume \(BA\) is a tangent, then \(\angle ABC=\frac{1}{2}\angle BOC\). Wait, no, the correct formula is: The measure of an inscribed angle \(\angle ABC\) subtended by an arc \(AC\) (assuming the standard circle - angle relation). The central angle \(\angle BOC\) and the inscribed angle \(\angle BAC\) subtend the same arc \(BC\). But if we consider the exterior angle of \(\triangle OBC\). Wait, no, the correct formula is:
The measure of an inscribed angle \(\angle ABC\) (assuming \(A\) is a point on the circumference such that \(\angle ABC\) is related to the arc \(AC\)). Wait, actually, using the property that the angle between a tangent and a chord is equal to the inscribed angle in the alternate segment. If \(BA\) is a tangent to the circle at \(B\), then \(\angle ABC=\frac{1}{2}\angle BOC\). Since \(\angle BOC = 70^{\circ}\), \(\angle ABC = 35^{\circ}\)

Answer:

\(35^{\circ}\)