Trigonometry (10th Edition)

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

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 315: 42

Answer

The diagonal opposite the angle of $58^{\circ}$ has a length of 5.2 cm The diagonal opposite the angle of $122^{\circ}$ has a length of 8.8 cm

Work Step by Step

Let $a = 4.0~cm$, let $b = 6.0~cm$, and let angle $C_1 = 58^{\circ}$. We can use the law of cosines to find $c_1$, the diagonal opposite the angle $C_1$: $c_1^2 = a^2+b^2-2ab~cos~C_1$ $c_1 = \sqrt{a^2+b^2-2ab~cos~C_1}$ $c_1 = \sqrt{(4.0~cm)^2+(6.0~cm)^2-(2)(4.0~cm)(6.0~cm)~cos~58^{\circ}}$ $c_1 = \sqrt{26.56~cm^2}$ $c_1 = 5.2~cm$ Let $a = 4.0~cm$, let $b = 6.0~cm$, and let angle $C_2 = 122^{\circ}$. We can use the law of cosines to find $c_2$, the diagonal opposite the angle $C_2$: $c_2^2 = a^2+b^2-2ab~cos~C_2$ $c_2 = \sqrt{a^2+b^2-2ab~cos~C_2}$ $c_2 = \sqrt{(4.0~cm)^2+(6.0~cm)^2-(2)(4.0~cm)(6.0~cm)~cos~122^{\circ}}$ $c_2 = \sqrt{77.44~cm^2}$ $c_2 = 8.8~cm$
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.