(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.
Explanation
(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°)