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°.