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: 20

Answer

$$A = {45^ \circ },\,\,B = {106.804^ \circ },\,\,\,c \approx 2.82{\text{in}}$$

Work Step by Step

$$\eqalign{ & C = {28.3^ \circ },\,\,\,b = 5.71{\text{in,}}\,\,\,a = 4.21{\text{in}} \cr & {\text{Use the Law of cosines to find }}c \cr & {c^2} = {a^2} + {b^2} - 2ab\cos C \cr & {\text{Substitute}} \cr & {c^2} = {\left( {4.21} \right)^2} + {\left( {5.71} \right)^2} - 2\left( {4.21} \right)\left( {5.71} \right)\cos \left( {{{28.3}^ \circ }} \right) \cr & {\text{Use a calculator}} \cr & {c^2} \approx 8{\text{in}} \cr & {\text{Take square roots and choose the positive root}} \cr & c \approx 2.8278{\text{in}} \cr & \cr & {\text{Calculate the angle }}A{\text{ using the law of sines}} \cr & \frac{a}{{\sin A}} = \frac{c}{{\sin C}} \cr & \sin A = \frac{{a\sin C}}{c} \cr & \sin A = \frac{{4.21\sin \left( {{{28.3}^ \circ }} \right)}}{{2.82}} \cr & A \approx {\sin ^{ - 1}}\left( {\frac{{4.21\sin \left( {{{28.3}^ \circ }} \right)}}{{2.8278}}} \right) \cr & A \approx {44.89^ \circ } \cr & \cr & {\text{Calculate }}B \cr & B = {180^ \circ } - A - C \cr & B = {180^ \circ } - {44.89^ \circ } - {28.3^ \circ } \cr & B = {106.804^ \circ } \cr} $$
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