Question

a ship leaves port a and travels 85 km due west to buoy c. it then adjusts its course 48° northward. it travels 129 km in that direction until it reaches port b. what angle θ with respect to due north could the ship have used to travel directly from port a to port b? see the figure below. (the figure is not drawn to scale.) carry your intermediate computations to at least four decimal places. round your answer to the nearest tenth of a degree. θ =

Explanation:

Step1: Find the angle at point C

The angle at point \( C\) is \(180^{\circ}- 48^{\circ}=132^{\circ}\)

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). In \(\triangle ABC\), let \(a = 129\), \(b\) be the side from \(A\) to \(B\), \(c = 85\), \(A=\theta\), \(B\) (angle opposite to side \(b\)) and \(C = 132^{\circ}\). So \(\frac{129}{\sin\theta}=\frac{85}{\sin(132^{\circ})}\)

$$ \sin\theta=\frac{129\times\sin(132^{\circ})}{85} $$

We know that \(\sin(132^{\circ})=\sin(180 - 48)^{\circ}=\sin48^{\circ}\approx0.7431\)

$$ \sin\theta=\frac{129\times0.7431}{85}=\frac{96.85}{85}\approx1.1394 $$

Wait, no, wrong. Wait, Law of Sines: \(\frac{BC}{\sin\theta}=\frac{AB}{\sin C}\) (no, correct formula is \(\frac{AC}{\sin B}=\frac{BC}{\sin A}=\frac{AB}{\sin C}\). Wait, in \(\triangle ABC\), \(AC = 85\), \(BC=129\), \(\angle C=132^{\circ}\). So \(\frac{129}{\sin\angle BAC}=\frac{85}{\sin\angle ABC}\). But we want \(\theta\), and \(\angle BAC\) and \(\theta\) are complementary (because of the north - west coordinate system).

First, use the Law of Cosines to find the length of \(AB\):

$$ AB^{2}=85^{2}+129^{2}-2\times85\times129\times\cos(132^{\circ}) $$

$$ \cos(132^{\circ})=\cos(180 - 48)^{\circ}=-\cos48^{\circ}\approx - 0.6691 $$

$$ AB^{2}=7225+16641-2\times85\times129\times(- 0.6691) $$

$$ AB^{2}=7225 + 16641+2\times85\times129\times0.6691 $$

$$ AB^{2}=23866+(2\times85\times129\times0.6691) $$

$$ 2\times85\times129 = 2\times10965=21930 $$

$$ 21930\times0.6691\approx14683.36 $$

$$ AB^{2}=23866 + 14683.36=38549.36\Rightarrow AB\approx196.34 $$

Now, use the Law of Sines: \(\frac{BC}{\sin\angle BAC}=\frac{AB}{\sin C}\)

$$ \sin\angle BAC=\frac{BC\times\sin C}{AB}=\frac{129\times\sin(132^{\circ})}{196.34} $$

$$ \sin(132^{\circ})\approx0.7431 $$

$$ \sin\angle BAC=\frac{129\times0.7431}{196.34}=\frac{96.86}{196.34}\approx0.4933 $$

\(\angle BAC\approx29.6^{\circ}\)

Since \(\theta\) and \(\angle BAC\) are complementary (because of the north - west direction, the angle between the north direction and the line \(AB\) and the angle between \(AB\) and the west direction at \(A\) add up to \(90^{\circ}\))
\(\theta = 90^{\circ}-\angle BAC\)

Another way:
The angle at \(C\) is \(132^{\circ}\), sides \(a = 85\), \(b = 129\)
By the Law of Sines: \(\frac{\sin\theta}{85}=\frac{\sin(132^{\circ})}{129}\)

$$ \sin\theta=\frac{85\times\sin(132^{\circ})}{129} $$

$$ \sin(132^{\circ})=\sin48^{\circ}\approx0.7431 $$

$$ \sin\theta=\frac{85\times0.7431}{129}=\frac{63.16}{129}\approx0.4896 $$

\(\theta\approx29.4^{\circ}\)

Answer:

\(29.4^{\circ}\)