Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 8 - Application of Trigonometry - 8.2 The Law of Cosines - 8.2 Exercises - Page 767: 36

Answer

$$A = {24.2^ \circ },\,\,\,\,B = {30.8^ \circ },\,\,\,C = {125^ \circ }$$

Work Step by Step

$$\eqalign{ & a = {\text{4}}.0{\text{ ft}},\,\,\,\,b = {\text{5}}.0{\text{ ft}},\,\,\,\,c = {\text{8}}.0{\text{ ft}} \cr & {\text{Use the law of cosines to solve for any angle of the triangle}} \cr & {c^2} = {a^2} + {b^2} - 2ab\cos C \cr & {\text{Solve for cos }}C \cr & \cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} \cr & {\text{Substitute}} \cr & \cos C = \frac{{{{\left( {{\text{4}}.0} \right)}^2} + {{\left( {{\text{5}}.0} \right)}^2} - {{\left( {{\text{8}}.0} \right)}^2}}}{{2\left( {{\text{4}}.0} \right)\left( {{\text{5}}.0} \right)}} \cr & {\text{Use a calculator}} \cr & \cos C = - \frac{{23}}{{40}} \cr & {\text{Use the inverse cosine function}} \cr & C \approx {125^ \circ } \cr & \cr & {\text{Use the Law of sines to find the angle of }}B \cr & \frac{{\sin B}}{b} = \frac{{\sin C}}{c} \cr & \sin B = \frac{{b\sin C}}{c} \cr & \sin B = \frac{{5.0\sin \left( {{{125}^ \circ }} \right)}}{{8.0}} \cr & {\text{Use a calculator}} \cr & \sin B \approx 0.5119700277 \cr & {\text{Use the inverse sine function}} \cr & B \approx {30.8^ \circ } \cr & \cr & {\text{Calculate }}A \cr & A = {180^ \circ } - B - C \cr & A = {180^ \circ } - {30.8^ \circ } - {125^ \circ } \cr & A = {30^ \circ } \cr & \cr & {\text{Answer}} \cr & A = {24.2^ \circ },\,\,\,\,B = {30.8^ \circ },\,\,\,C = {125^ \circ } \cr} $$
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