(a) Given: Circle ABC with centre O To prove : < AOB = 2 < ACB Construction: Join CO produced to P. Proof: With lettering as in the figure, OA = OB (radii) ; \(x_{1} = x_{2}\) (base angles of an isosceles triangle) \(\therefore < AOP = x_{1} + x_{2}\) (exterior angle of triangle AOC) \(\therefore < AOP = 2x_{2} (x_{1} = x_{2})\) Also, \(< BOP = 2y_{2}\) (similar proof as the earlier done ones) \(\therefore < AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2})\) \(< AOB = 2 \times < ABC\) (proven) (b) From the diagram, < OQR = 32^\circ and < MPQ = 15^\circ. (i) \(\therefore < ORQ = 32^\circ\) (base angles of an isosceles triangle) \(< OQR = 180^\circ - (32^\circ + 32^\circ) = 180^\circ - 64^\circ = 116^\circ\) \(\therefore < QPR = \frac{116^\circ}{2} = 58^\circ\) (angle subtended at the centre) (ii) Join MO and MQ, Since < MPQ = 15^\circ then < MOQ = 2 < MPQ (angle at the centre) \(\therefore < MOQ = 2 \times 15^\circ = 30^\circ\) \(\therefore < OMQ = < MQO \) (base angles of an isosceles triangle) \(\therefore < MQO = \frac{180^\circ - 30^\circ}{2} = 75^\circ\)