(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, < OQR = 32° and < MPQ = 15°. Calculate (i) < QPR ; (ii) < MQO.
Explanation
(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° and < MPQ = 15°.
(i) (therefore < ORQ = 32°) (base angles of an isosceles triangle)
(< OQR = 180° – (32° + 32°) = 180° – 64° = 116°)
(therefore < QPR = frac{116°}{2} = 58°) (angle subtended at the centre)
(ii) Join MO and MQ,
Since < MPQ = 15°
then < MOQ = 2 < MPQ (angle at the centre)
(therefore < MOQ = 2 times 15° = 30°)
(therefore < OMQ = < MQO ) (base angles of an isosceles triangle)
(therefore < MQO = frac{180° – 30°}{2} = 75°)