## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Review Exercises - Page 352: 52

#### Answer

The resulting speed of the boat is 20.85 km/h The final bearing of the boat is $117.5^{\circ}$

#### Work Step by Step

Let $a = 15~km/h$ Let $b = 7~km/h$ Let angle $\theta$ be the angle between these two vectors. Then $\theta = 130^{\circ}-90^{\circ} = 40^{\circ}$ Let $c$ be the resultant of these two vectors. Note that the magnitude of $c$ is the resulting speed. We can use the parallelogram rule to find $c$: $c = \sqrt{a^2+b^2+2ab~cos~\theta}$ $c = \sqrt{(15~km/h)^2+(7~km/h)^2+(2)(15~km/h)(7~km/h)~cos~40^{\circ}}$ $c = \sqrt{434.87~(km/h)^2}$ $c = 20.85~km/h$ The resulting speed of the boat is 20.85 km/h We can use the law of sines to find the angle A between the vectors b and c: $\frac{c}{sin~C} = \frac{a}{sin~A}$ $sin~A = \frac{a~sin~C}{c}$ $A = arcsin(\frac{a~sin~C}{c})$ $A = arcsin(\frac{(15)~sin~140^{\circ}}{20.85})$ $A = arcsin(0.46244)$ $A = 27.5^{\circ}$ The final bearing of the boat is $90^{\circ}+27.5^{\circ} = 117.5^{\circ}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.