Question

A ship travels from Akron (A) on a bearing of 030° to Bellvilie (B),90km away. It then travels to Comptin (C) which is 310 km due east of Akron (A).Draw a diagram showing the movement of the ship indicate:a.)Northb.)the distances travelledc.)the know angles

Answer 1

Given: The bearing and distance of a ship from Akron(A) through Bellville(B) to Compton(C)

To Determine: The distance travelled and the angles

Solution

The diagram of the journey can be represented as shown below

Let us solve for BC using cosine rule

$\begin{gathered} BC^2=AB^2+AC^2-2(AB)(AC)cosA \\ BC^2=90^2+310^2-2(90)(310)(cos60^0) \\ BC^2=8100+96100-27900 \\ BC^2=76300 \\ BC=\sqrt{76300} \\ BC=276.22km \\ BC\approx276km \end{gathered}$

The total distance travelled is

$\begin{gathered} Total-distance=AB+BC+AC \\ =90km+276km+310km \\ =676km \end{gathered}$

Using sine rule, we can determine the measure of angle C

$\begin{gathered} \frac{AB}{sinC}=\frac{BC}{sinA}=\frac{AC}{sinB} \\ \frac{90}{sinC}=\frac{276}{sin60^0}=\frac{310}{sinB} \\ \frac{90}{sinC}=\frac{276}{sin60^0} \\ sinC=\frac{90(sin60^0)}{276} \\ sinC=\frac{155.88}{276} \\ sinC=0.5648 \\ C=sin^{-1}(0.5648) \\ C=34.39^0 \end{gathered}$

Also know that the sum of angles in a triangle is 180 degree. Therefore

$\begin{gathered} A+B+C=180^0 \\ 60^0+B+34.39^0=180^0 \\ B+94.39^0=180^0 \\ B=180^0-94.39^0 \\ B=85.61^0 \\ B\approx85.6^0 \end{gathered}$

The bearing of Compton from Bellville(B) is

$Bearing(C-from-B)=90^0+34.39^0=124.39^0\approx124.4^0$

The bearring of Akron(A) from Compton(C) is

$Bearing(A-from-C)=270^0+34.39^0=304.39^0\approx304.4^0$

In summary

The total distance travelled is 676km

The bearing of Compton from Bellville is 124.4 degrees, and the bearing of Akron from Compton is 304.4 degrees