\(A\), \(B\), \(C\) and \(D\) are points on a circle and \(AB\) is a diameter. Angle \(BAC = 29^\circ\) and angle \(CBD = 43^\circ\). (a) Write down angle \...

The diameter \(AB\) drives most of this question, because any angle subtended by a diameter at a point on the circle is a right angle. The rest uses the cyclic quadrilateral property and the same-segment property.

1. (a) Angle \(ACB\). \(AB\) is a diameter and \(C\) is on the circle, so

   angle \(ACB = 90^\circ\) [B1], because the angle in a semicircle is a right angle [B1]

2. (b) Angle \(ABC\). The angles of triangle \(ABC\) add to \(180^\circ\), with \(90^\circ\) at \(C\) and \(29^\circ\) at \(A\):

   angle \(ABC = 180 - 90 - 29 = 61^\circ\) [B1]

3. (c) Angle \(ADC\). \(ABCD\) is a cyclic quadrilateral, and \(B\) and \(D\) are opposite vertices, so angles \(ABC\) and \(ADC\) are supplementary:

   angle \(ADC = 180 - 61 = 119^\circ\) [M1] [A1]

   Reason: opposite angles of a cyclic quadrilateral add to \(180^\circ\).

4. (d) Angle \(CAD\). Angles \(CAD\) and \(CBD\) both stand on the chord \(CD\), with \(A\) and \(B\) on the same side of it, so they are angles in the same segment and are equal:

   angle \(CAD = 43^\circ\) [B1], angles in the same segment standing on arc \(CD\) [B1]

5. (e) Angle \(ADB\). \(AB\) is a diameter and \(D\) is on the circle, so the angle in a semicircle applies again:

   angle \(ADB = 90^\circ\) [B1]

Check the quadrilateral: angle \(BAD = 29 + 43 = 72^\circ\) at \(A\), and \(72 + 108 = 180^\circ\) requires angle \(BCD = 108^\circ\), which is consistent with angle \(BCA = 90^\circ\) plus angle \(ACD = 180 - 119 - 43 = 18^\circ\). Part (d) is where the same-segment theorem is easily confused with the centre-circumference theorem; the test is that both angles sit on the circumference and stand on the same chord, which \(CAD\) and \(CBD\) do.