Chapter 6.2, Problem 51E

To determine

To find:Draw the diagram that gives a visual representation of the given information. Then, find the bearings for the last two legs of the race.

Answer to Problem 51E

The bearing for the second and third legs of the race are $\text{N}37.1°\text{E and S}63.1°\text{E}$ , respectively.

Explanation of Solution

Given information:

A boat race runs along a triangular course marked by buoys $A,B,\text{and }C$ . The race starts with the boats headed west for 3700 meters. The other two sides of the course lie to the north of the first side, and their lengths are 1700 meters and 3000 meters.

Calculations:

Drawing the given scenario,

  

To find $\angle B$ :

  $\begin{array}{l}{b}^2=a^2+c^2-2ac\cos (B)\\ \Rightarrow \cos (B)=\frac{a^2+c^2-b^2}{2ac}\\ \Rightarrow \cos (B)=\frac{{1700}^2+{3700}^2-{3000}^2}{2(1700)(3700)}\\ \Rightarrow B=52.9°\end{array}$

To find $\angle C$ :

  $\begin{array}{l}{c}^2=a^2+b^2-2ab\cos (C)\\ \Rightarrow \cos (C)=\frac{a^2+b^2-c^2}{2ab}\\ \Rightarrow \cos (C)=\frac{{1700}^2+{3000}^2-{3700}^2}{2(1700)(3000)}\\ \Rightarrow C=100.2°\end{array}$

Conclusion:

Therefore, the bearing of the second leg of the race is

  $90°-52.9°=\text{N}37.1°\text{E}$ .

And that of the third leg of the race is

  $100.2°-37.1°=\text{S}63.1°\text{E}$ .