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: 17

Answer

$$b = 88.2,{\text{ }}A = {56.7^ \circ },\,\,C = {68.3^ \circ }$$

Work Step by Step

$$\eqalign{ & {\text{From the triangle we have:}} \cr & B = {55^ \circ },{\text{ }}a = 90\,{\text{ and }}\,c = 100{\text{ }} \cr & \cr & {\text{Use the Law of cosines to find }}b \cr & {b^2} = {a^2} + {c^2} - 2ac\cos B \cr & {\text{Substitute}} \cr & {b^2} = {\left( {90} \right)^2} + {\left( {100} \right)^2} - 2\left( {90} \right)\left( {100} \right)\cos \left( {{{55}^ \circ }} \right) \cr & {\text{Use a calculator}} \cr & {b^2} \approx 7775.624146 \cr & {\text{Take square roots and choose the positive root}} \cr & b \approx 88.2 \cr & \cr & {\text{Calculate the angle }}A{\text{ using the law of sines}} \cr & \frac{{\sin A}}{a} = \frac{{\sin B}}{b} \cr & \sin A = \frac{{a\sin B}}{b} \cr & \sin A = \frac{{90\sin \left( {{{55}^ \circ }} \right)}}{{88.2}} \cr & {\text{Use a calculator}} \cr & \sin A \approx 0.8358694329 \cr & {\text{Use the inverse sine function}} \cr & A \approx {56.7^ \circ } \cr & \cr & {\text{Calculate }}C \cr & C = {180^ \circ } - A - B \cr & C = {180^ \circ } - {56.7^ \circ } - {55^ \circ } \cr & C = {68.3^ \circ } \cr & \cr & {\text{Answer}} \cr & b = 88.2,{\text{ }}A = {56.7^ \circ },\,\,C = {68.3^ \circ } \cr} $$
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