In the diagram, /PQ/ = 8m, /QR/ = 13m, the bearing of Q from P is 050° and the bearing of R from Q is 130°.
(a) Calculate, correct to 3 significant figures, (i) /PR/ ; (ii) the bearing of R from P.
(b) Calculate the shortest distance between Q and PR, hence the area of triangle PQR.
Explanation
(a)(i) < PQR = (180° – 130°) + 50° = 100°
(PR^{2} = PQ^{2} + QR^{2} – 2(PQ)(QR) cos < PQR)
= (8^{2} + 13^{2} – 2(8)(13) cos 100)
= (233 + 36.118)
(PR^{2} = 269.118)
(PR = 16.405 m approxeq 16.4 m)
(ii) (frac{QR}{sin < QPR} = frac{PR}{sin < PQR})
(frac{13}{sin < QPR} = frac{16.4}{sin 100})
(sin < QPR = frac{13 times sin 100}{16.4})
(sin < QPR = 0.7806)
(< QPR = 51.32° approxeq 51.3°)
The bearing of R from P = 50° + 51.3° = 101.3°
(approxeq) 101°.
(b)
In (Delta PQD),
(frac{h}{8} = sin 51.3)
(h = 8 sin 51.3)
= (6.243 m )
(approxeq 6.24 m)
Area of (Delta PQR = frac{1}{2} times PR times h)
(frac{1}{2} times 6.24 times 16.4 = 51.166 m^{2})
(approxeq 51.2 m^{2})