Precalculus (6th Edition)

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

Chapter 8 - Application of Trigonometry - 8.1 The Law of Sines - 8.1 Exercises - Page 754: 34

Answer

$${A_1} = {72.16^ \circ },\,\,\,\,\,\,{A_2} = {107.84^ \circ },\,\,\,\,\,\,\,{C_1} = {59.64^ \circ },\,\,\,\,{C_2} = {23.96^ \circ }$$

Work Step by Step

$$\eqalign{ & B = {48.2^ \circ },\,\,\,a = 890{\text{cm,}}\,\,\,b = 697{\text{cm}} \cr & \cr & {\text{Use the law of sines to find the angle of }}A \cr & \frac{a}{{\sin A}} = \frac{b}{{\sin B}} \cr & \sin A = \frac{{a\sin B}}{b} \cr & \sin A = \frac{{890\sin \left( {{{48.2}^ \circ }} \right)}}{{697}} \cr & {\text{Use a calculator}} \cr & \sin A = 0.951899 \cr & \cr & {\text{There are two }}\,{\text{angles }}B\,{\text{between }}\,{{\text{0}}^ \circ }{\text{ and 18}}{{\text{0}}^ \circ }{\text{ that satisfy this }} \cr & {\text{condition}}{\text{.}} \cr & {A_1} = {\sin ^{ - 1}}\left( {0.951899} \right) \cr & {\text{Use the inverse sine function}} \cr & {A_1} = {72.16^ \circ } \cr & \cr & {\text{Supplementary angles have the same sine value}},{\text{ so another }} \cr & {\text{possible value of angle }}A{\text{ is}} \cr & {A_2} = {180^ \circ } - {72.16^ \circ } \cr & {A_2} = {107.84^ \circ } \cr & \cr & {\text{Calculating }}{C_1} \cr & {C_1} = {180^ \circ } - B - {A_1} \cr & {C_1} = {180^ \circ } - {48.2^ \circ } - {72.16^ \circ } \cr & {C_1} = {59.64^ \circ } \cr & \cr & {\text{Calculating }}{C_2} \cr & {C_2} = {180^ \circ } - B - {A_2} \cr & {C_2} = {180^ \circ } - {48.2^ \circ } - {107.84^ \circ } \cr & {C_2} = {23.96^ \circ } \cr & \cr & {\text{Answer}} \cr & {A_1} = {72.16^ \circ },\,\,\,\,\,\,{A_2} = {107.84^ \circ },\,\,\,\,\,\,\,{C_1} = {59.64^ \circ },\,\,\,\,{C_2} = {23.96^ \circ } \cr} $$
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