(a) Given : Circle ABC, centre O. To prove: < AOB = < ACB Construction : Join CO produced to P. Proof: With the lettering in the figure, OA = OB (radii) \(x_{1} = x_{2}\) (base angles of isosceles triangle) \(\therefore < AOP = x_{1} + x_{2}\) (exterior angle of triangle AOC) \(\therefore < AOP = 2x_{2} (x_{1} = x_{2})\) (base angles of an isosceles triangle) Similarly, \(< BOP = 2y_{2}\) \(\therefore < AOB = 2x_{2} + 2y_{2} \) = \(2(x_{2} + y_{2})\) \(\implies < AOB = 2 \times < ACB\) (proved) (b) From the figure, \(< ACB = \frac{130^\circ}{2} = 65^\circ\) (angle subtended at the centre of the circle) \(< ACO = 26^\circ\) (Base angles of isosceles triangle ACO) \(\therefore < BCO = 65^\circ - 26^\circ = 39^\circ\) \(< OBC = 39^\circ\) (base angles of isosceles triangle OBC) \(\therefore < COB = 180^\circ - (39^\circ + 39^\circ) = 102^\circ\)