(a) (i) 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.
(ii) In the diagram above, O is the centre of the circle and PT is a diameter. If < PTQ = 22^\circ and < TOR = 98^\circ, calculate < QRS.
(b) ABCD is a cyclic quadrilateral and the diagonals AC and BD intersect at H. If < DAC = 41^\circ and < AHB = 70^\circ, calculate < ABC.
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a)(i)
Given a circle APB with centre O.
To prove that < AOB = 2 < APB.
Construction: Draw PO and produce it to Q.
Proof : Referring to the lettering in the figure, |OA| = |OB| (radius)
\(\therefore x_{1} = x_{2}\) (base angles of isosceles triangle)
\(< AOQ = x_{1} + x_{2}\) (exterior angles of triangle AOP)
\(\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\)
Similarly, \(< BOQ = 2y_{2}\)
\(< AOB = < AOQ + < BOQ \)
Hence, \(< AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2})\)
= \(2 \times < APB\)
(ii)
From the figure, \(|OQ| = |QT|\) (radius)
\(< OQT = 22^\circ = < QTO\) (base angles of isosceles triangle)
\(\therefore < POQ = 44^\circ \) (exterior angle of \(\Delta\) QOT)
\(< QOR = 180^\circ - (44^\circ + 98^\circ) = 180^\circ - 142^\circ = 38^\circ\) (Angles on a straight line)
\(< QOR = 2 \times < QSR = 38^\circ\)
\(\therefore < QSR = \frac{38^\circ}{2} = 19^\circ\)
(b)
\(< BHC = (180^\circ - 70^\circ) = 110^\circ\) (angle on a straight line)
\(< DAC = 41^\circ = < DBC\) (angles on the same segment)
But \(< DBC = < HBC\)
\(< AHB = < HBC + < HCB \) (exterior angles of \(\Delta\) BHC)
\(70^\circ = 41^\circ + < HCB \implies < HCB = 70^\circ - 41^\circ = 29^\circ\)
\(< ACB = < HCB = 29^\circ\)
Source: Myschool Past Questions Archive (Record #54687)