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 nearest degree. < MON.
(b) T is equidistant from P and Q. The bearing of P from T is 60(^o) and the bearing of Q from T is 130(^o).
(i) Illustrate the information on a diagram.
(ii) Find the bearing of Q from P.
Explanation
(a) Using length of chord = 2r sin ((frac{theta}{2})) = 22.42 and simplifying sin((frac{theta}{2})) = (frac{22.42}{33.18}) = 0.6757
Taking the sine inverse of 0.6757 to obtain (frac{theta}{2}) = 42.51(^o) and multiplying through to get (theta) = 58.02(^o) = 885(^o)
(b(i), the expected diagram is
(b)(ii) since T is equidistant from P and Q, then ITPI = ITQI. Using sum of the angles in a triangle. 70° + x + x = 180° and solving for x.
x = 55°. Therefore. the bearing of Q from P = 270° — 30° — 55° = 185°.