Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 323: 59


The length of the tunnel is 5512.5 m.

Work Step by Step

Note that the angle at point $C$ is $110^{\circ}$. We can use the law of cosines to find the length $AB$, which is the length of the tunnel: $AB^2 = AC^2+BC^2-2(AC)(BC)~cos~C$ $AB = \sqrt{AC^2+BC^2-2(AC)(BC)~cos~C}$ $AB = \sqrt{(3800~m)^2+(2900~m)^2-(2)(3800~m)(2900~m)~cos~110^{\circ}}$ $AB = \sqrt{30388123.96~m^2}$ $AB = 5512.5~m$ The length of the tunnel is 5512.5 m
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.