Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.5 Applications of Vectors - 7.5 Exercises - Page 336: 16

Answer

The liner is a distance of 6.6 miles from the port. The bearing from port is $117.1^{\circ}$

Work Step by Step

Let $a = 8.8~mi$ Let $b = 2.4~mi$ Let angle $C$ be the angle between these two vectors. Then $C = 90^{\circ}-70^{\circ} = 20^{\circ}$ We can use the law of cosines to find $c$, the distance of the liner from the port: $c = \sqrt{a^2+b^2-2ab~cos~C}$ $c = \sqrt{(8.8~mi)^2+(2.4~mi)^2-(2)(8.8~mi)(2.4~mi)~cos~20^{\circ}}$ $c = \sqrt{43.5~mi^2}$ $c = 6.6~mi$ The liner is a distance of 6.6 miles from the port. We can use the law of sines to find the angle $B$ between the resultant vector and the vector of $8.8 ~mi$: $\frac{c}{sin~C} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~C}{c}$ $B = arcsin(\frac{b~sin~C}{c})$ $B = arcsin(\frac{(2.4~mi)~sin~20^{\circ}}{6.6~mi})$ $B = arcsin(0.12437)$ $B = 7.1^{\circ}$ The bearing from port is $110^{\circ}+7.1^{\circ}$ which is $117.1^{\circ}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.