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 320: 30


The angles of the triangle are as follows: $A = 63^{\circ}50', B = 43^{\circ}30',$ and $C = 72^{\circ}40'$ The lengths of the sides are as follows: $a = 327~ft, b = 251~ft,$ and $c = 348~ft$

Work Step by Step

We can use the law of cosines to find $c$: $c^2 = a^2+b^2-2ab~cos~C$ $c = \sqrt{a^2+b^2-2ab~cos~C}$ $c = \sqrt{(327~ft)^2+(251~ft)^2-(2)(327~ft)(251~ft)~cos~72^{\circ}40'}$ $c = \sqrt{121024.46~ft^2}$ $c = 348~ft$ We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{(327)^2+(348)^2-(251)^2}{(2)(327)(348)})$ $B = arccos(0.725)$ $B = 43^{\circ}30'$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-43^{\circ}30'-72^{\circ}40'$ $A = 63^{\circ}50'$
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.