NOT DRAWN TO SCALE In the diagram, is a chord of a circle with centre 0. 22.42 cm and the perimeter of triangle MON is 55.6 cm. Calculate, correct to the ne...

(a) Angle MON

From the diagram, MN is a chord of the circle with centre O, and \(|MN| = 22.42\ \text{cm}\). OM and ON are radii, so \(|OM| = |ON| = r\).

The perimeter of triangle MON is 55.6 cm:

\[|OM| + |ON| + |MN| = 55.6\]

\[r + r + 22.42 = 55.6\]

\[2r = 33.18 \quad\Rightarrow\quad r = 16.59\ \text{cm}\]

Triangle MON is isosceles with \(OM = ON = 16.59\). Drop the perpendicular from O to the midpoint of MN; it bisects angle MON. Then

\[\sin\!\left(\frac{\angle MON}{2}\right) = \frac{\tfrac{1}{2}|MN|}{r} = \frac{11.21}{16.59} = 0.6757\]

\[\frac{\angle MON}{2} = \sin^{-1}(0.6757) = 42.5^\circ\]

\[\angle MON = 85.0^\circ \approx 85^\circ\]

(b) Bearings

(i) Illustration. T is equidistant from P and Q, so \(|TP| = |TQ|\). From T, the bearing of P is \(060^\circ\) and the bearing of Q is \(130^\circ\). Measuring both angles clockwise from the north line at T, P lies to the north-east and Q further round to the south-east.

The angle between the two bearings at T is

\[\angle PTQ = 130^\circ - 60^\circ = 70^\circ\]

(ii) Bearing of Q from P. Since \(|TP| = |TQ|\), triangle PTQ is isosceles, so the base angles are equal:

\[\angle TPQ = \angle TQP = \frac{180^\circ - 70^\circ}{2} = 55^\circ\]

The bearing of T from P is the reverse of the bearing of P from T:

\[060^\circ + 180^\circ = 240^\circ\]

Q lies \(55^\circ\) from PT, measured back towards the south. Hence the bearing of Q from P is

\[240^\circ - 55^\circ = 185^\circ\]

The bearing of Q from P is \(\mathbf{185^\circ}\).