(a) Triangle PQR is right-angled at Q. PQ = 3a cm and QR = 4a cm. Determine PR in terms of a.
(b) Ayo travels a distance of 24km from X on a bearing of 060° to Y. He then travels a distance of 18km to a point Z and Z is 30km from X.
(i) Draw the diagram to show the positions of X, Y and Z ; (ii) What is the bearing of Z from Y ; (iii) Calculate the bearing of X from Z.
Explanation
(a)
From the diagram and by Pythagoras theorem,
(|PR|^{2} = |PQ|^{2} + |QR|^{2})
(|PR|^{2} = (3a)^{2} + (4a)^{2})
(|PR|^{2} = 9a^{2} + 16a^{2} = 25a^{2})
(|PR| = sqrt{25a^{2}} = 5a)
(b)
(ii) From the figure, Bearing of Y from X = 060°
Using cosine rule,
(cos XYZ = frac{24^{2} + 18^{2} – 30^{2}}{2 times 24 times 18})
(cos XYZ = 0 implies < XYZ = 90°)
(< RYZ = 90° – 60° = 30°)
(< N_{2} YZ = 180° – 30° = 150°) (angles on a straight line)
The bearing of Z from Y is 150°.
(iii) (cos XZY = frac{30^{2} + 18^{2} – 24^{2}}{2 times 30 times 18})
= (frac{900 + 324 – 576}{1080})
(cos XZY = frac{648}{1080} = 0.6)
(< XZY = cos^{-1} (0.6) = 53.13°)
(therefore) The bearing of X from Y = 360° – (53.13° + 30°) = 276.87°.