Answer: (b) \(m\angle CAT = 90^\circ\); (c) \(m\angle ABC = 90^\circ\).
(a) Figure (as drawn): draw a circle with centre O. Draw diameter AC (A, O, C collinear). Take a point B on the circle (on either arc AC) and draw chord AB and seg BC. At A draw line TA perpendicular to OA (point T on the line outside the circle); line AT is the tangent at A. Mark the right angle at A between AT and AC, and the angle at B.
(b) Line AT is a tangent at A and seg OA is the radius through the point of contact, so by the tangent theorem AT \(\perp\) OA, i.e. \(m\angle OAT = 90^\circ\). Since A, O, C are collinear, \(\angle CAT\) and \(\angle OAT\) are the same angle.
\(\therefore m\angle CAT = 90^\circ\).
(c) \(\angle ABC\) is inscribed in arc ABC and intercepts arc AC, which is a semicircle (AC is a diameter), so \(m(\text{arc AC}) = 180^\circ\).
\(\therefore m\angle ABC = \dfrac{1}{2} \times 180^\circ = 90^\circ\) (angle inscribed in a semicircle is a right angle).
(Note: a picture cannot be shown here; in the examination the figure must actually be drawn and labelled as described.)
