## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 320: 37

#### Answer

When we try to use the law of cosines to find any angle A, B, or C, the angles are not defined. This makes sense since no triangle exists with the given values in the question.

#### Work Step by Step

We can try to use the law of cosines to find $A$: $a^2 = b^2+c^2-2bc~cos~A$ $2bc~cos~A = b^2+c^2-a^2$ $cos~A = \frac{b^2+c^2-a^2}{2bc}$ $cos~A = \frac{4^2+10^2-3^2}{(2)(4)(10)}$ $cos~A = 1.3375$ Since there is no angle $A$ such that $cos~A \gt 1$, the angle $A$ is not defined. We can try to use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $cos~B = \frac{3^2+10^2-4^2}{(2)(3)(10)}$ $cos~B = 1.55$ Since there is no angle $B$ such that $cos~B \gt 1$, the angle $B$ is not defined. We can try to use the law of cosines to find $C$: $c^2 = a^2+b^2-2ab~cos~C$ $2ab~cos~C = a^2+b^2-c^2$ $cos~C = \frac{a^2+b^2-c^2}{2ab}$ $cos~C = \frac{3^2+4^2-10^2}{(2)(3)(4)}$ $cos~C = -3.125$ Since there is no angle $C$ such that $cos~C \lt -1$, the angle $C$ is not defined.

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.