(a) Given that \(5 \cos (x + 8.5)° - 1 = 0, 0° \leq x \leq 90°\), calculate, correct to the nearest degree, the value of x. (b) The bearing of Q from P is 0...

Answer Details

(a) Solving \(5\cos(x+8.5)^\circ - 1 = 0\).

\[\cos(x+8.5)^\circ = \frac{1}{5} = 0.2\]

\[x + 8.5 = \cos^{-1}(0.2) = 78.46^\circ\]

\[x = 78.46 - 8.5 = 69.96^\circ \approx 70^\circ\]

(b) Reading the bearings: the bearing of \(Q\) from \(P\) is \(150^\circ\) and the bearing of \(P\) from \(R\) is \(015^\circ\). Since the bearing of \(P\) from \(R\) is \(015^\circ\), the bearing of \(R\) from \(P\) is \(015^\circ + 180^\circ = 195^\circ\). With \(|PQ| = 24\) km and \(|PR| = 32\) km:

(i) Diagram. At \(P\), draw the north line. \(Q\) lies on a bearing of \(150^\circ\) at 24 km; \(R\) lies on a bearing of \(195^\circ\) at 32 km. The angle \(\angle QPR = 195^\circ - 150^\circ = 45^\circ\).

(ii) Distance \(|QR|\). By the cosine rule in \(\triangle PQR\):

\[|QR|^2 = 24^2 + 32^2 - 2(24)(32)\cos 45^\circ\]

\[= 576 + 1024 - 1536(0.7071) = 1600 - 1086.11 = 513.89\]

\[|QR| = \sqrt{513.89} = 22.67\text{ km (2 d.p.)}\]

(iii) Bearing of \(R\) from \(Q\). Taking \(P\) as origin with north as the \(y\)-axis: \(Q = (24\sin150^\circ,\,24\cos150^\circ) = (12,\,-20.78)\) and \(R = (32\sin195^\circ,\,32\cos195^\circ) = (-8.28,\,-30.91)\).

Displacement from \(Q\) to \(R\): \((-20.28,\,-10.13)\), which points south and west (third quadrant).

\[\theta = \tan^{-1}\!\left(\frac{20.28}{10.13}\right) = 63.5^\circ \text{ west of south}\]

\[\text{Bearing of }R\text{ from }Q = 180^\circ + 63.5^\circ = 243.5^\circ \approx 243^\circ\]