Question

1. surveying a surveyor wants to find the distance across a swamp (see figure). the bearing from a to b is n 32° w. the surveyor walks 50 meters from a, and at the point c the bearing to b is n 68° w. find (a) the bearing from a to c and (b) the distance from a to b.

Explanation:

Step1: Find the bearing from \(A\) to \(C\)

Since the bearing from \(A\) to \(B\) is \(N32^{\circ}W\) and the bearing from \(C\) to \(B\) is \(N68^{\circ}W\). The bearing from \(A\) to \(C\) is \(S90^{\circ}-32^{\circ} = S58^{\circ}E\) (by the property of angles in direction - bearing relations).

Step2: Find angles in \(\triangle ABC\)

Let \(\angle BAC=\alpha = 32^{\circ}\), \(\angle ACB=\beta=68^{\circ}- 32^{\circ}=36^{\circ}\), then \(\angle ABC = 180^{\circ}-(32^{\circ}+36^{\circ}) = 112^{\circ}\)
We know \(AC = 50\) meters. Using the Law of Sines \(\frac{AB}{\sin\beta}=\frac{AC}{\sin\angle ABC}\)

Step3: Calculate \(AB\)

Substitute \(\beta = 36^{\circ}\), \(AC = 50\), \(\angle ABC=112^{\circ}\) into the Law of Sines formula \(AB=\frac{AC\sin\beta}{\sin\angle ABC}\)
\(AB=\frac{50\times\sin36^{\circ}}{\sin112^{\circ}}\)
Since \(\sin36^{\circ}\approx0.5878\), \(\sin112^{\circ}=\sin(90^{\circ} + 22^{\circ})=\cos22^{\circ}\approx0.9272\)
\(AB=\frac{50\times0.5878}{0.9272}\approx31.7\) meters

Answer:

(a) The bearing from \(A\) to \(C\) is \(S58^{\circ}E\)
(b) The distance from \(A\) to \(B\) is approximately \(31.7\) meters