(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 circumfer...

Answer Details

(a) Proof: the angle at the centre is twice the angle at the circumference

Given: A circle centre \(O\), an arc \(AB\), with \(\angle AOB\) at the centre and \(\angle ACB\) at a point \(C\) on the remaining part of the circumference, both standing on the same arc \(AB\).

To prove: \(\angle AOB = 2\,\angle ACB\).

Construction: Join \(C\) to \(O\) and produce it to a point \(D\).

Proof: In triangle \(OAC\), \(OA = OC\) (radii), so it is isosceles and \(\angle OAC = \angle OCA\). The exterior angle equals the sum of the two opposite interior angles:

\[ \angle AOD = \angle OAC + \angle OCA = 2\,\angle OCA \]

Similarly, in triangle \(OBC\), \(OB = OC\) (radii), so \(\angle OBC = \angle OCB\) and

\[ \angle BOD = \angle OBC + \angle OCB = 2\,\angle OCB \]

Adding:

\[ \angle AOD + \angle BOD = 2\,\angle OCA + 2\,\angle OCB \]

\[ \angle AOB = 2(\angle OCA + \angle OCB) = 2\,\angle ACB \]

Hence the angle subtended at the centre is twice that subtended at the circumference. (The same argument holds when \(O\) lies outside triangle \(ACB\), using subtraction instead of addition.) \(\blacksquare\)

(b) Calculations

\(O\) is the centre, \(\angle OQR = 32^{\circ}\) and \(\angle MPQ = 15^{\circ}\).

(i) \(\angle QPR\)

In triangle \(OQR\), \(OQ = OR\) (radii), so it is isosceles and \(\angle ORQ = \angle OQR = 32^{\circ}\). Then

\[ \angle QOR = 180^{\circ} - 32^{\circ} - 32^{\circ} = 116^{\circ} \]

\(\angle QOR\) is the angle at the centre and \(\angle QPR\) is the angle at the circumference standing on the same arc \(QR\). By the theorem proved in (a):

\[ \angle QPR = \tfrac{1}{2}\,\angle QOR = \tfrac{1}{2}\times 116^{\circ} = 58^{\circ} \]

(ii) \(\angle MQO\)

\(\angle MPQ = 15^{\circ}\) is the angle at the circumference standing on arc \(MQ\). The angle at the centre on the same arc is

\[ \angle MOQ = 2 \times 15^{\circ} = 30^{\circ} \]

In triangle \(OMQ\), \(OM = OQ\) (radii), so it is isosceles with equal base angles:

\[ \angle MQO = \angle OMQ = \frac{180^{\circ} - 30^{\circ}}{2} = 75^{\circ} \]

\(\angle QPR = 58^{\circ}\) and \(\angle MQO = 75^{\circ}\).