In the diagram, /AB/ = 8 km, /BC/ = 13 km, the bearing of A from B is 310° and the bearing of B from C is 230°. Calculate, correct to 3 significant figures,...

Answer Details

Setting up the angle at B.

The bearing of A from B is \(310°\), so \(BA\) lies \(360°-310°=50°\) west of north at B (the diagram marks \(40°\) between the west arm and \(BA\), and \(90°-50°=40°\)).

The bearing of B from C is \(230°\), so the bearing of C from B is \(230°-180°=050°\); thus \(BC\) lies \(50°\) east of north at B (marked \(50°\) at C).

\[\angle ABC = 50°+50° = 100°.\]

(a) Distance AC. By the cosine rule,

\[AC^2 = AB^2+BC^2-2\,AB\cdot BC\cos\angle ABC\]

\[AC^2 = 8^2+13^2-2(8)(13)\cos100° = 233-208(-0.1736)=269.1\]

\[AC=\sqrt{269.1}=16.4\text{ km (3 s.f.)}\]

(b) Bearing of C from A. By the sine rule,

\[\frac{\sin\angle BAC}{BC}=\frac{\sin\angle ABC}{AC}\Rightarrow \sin\angle BAC=\frac{13\sin100°}{16.4}=0.7804\]

\[\angle BAC=51.3°.\]

The bearing of B from A is \(310°-180°=130°\). C lies to the north of the line AB, nearer A's north arm, so the bearing of C from A is

\[130°-51.3°=078.7°\approx 079°.\]

(c) How far east of B is C. C is on bearing \(050°\) from B at a distance of \(13\) km, so the eastward component is

\[13\sin50°=13(0.7660)=9.96\text{ km (3 s.f.)}\]