(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.
(b) 
In the diagram, O is the centre of the circle ACDB. If < CAO = 26° and < AOB = 130°. Calculate : (i) < OBC ; (ii) < COB.
(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, \(< ACB = \frac{130°}{2} = 65°\) (angle subtended at the centre of the circle)
\(< ACO = 26°\) (Base angles of isosceles triangle ACO)
\(\therefore < BCO = 65° - 26° = 39°\)
\(< OBC = 39°\) (base angles of isosceles triangle OBC)
\(\therefore < COB = 180° - (39° + 39°) = 102°\)
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