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: 22

Answer

$$A = {49^ \circ }40',\,\,\,b = 16.1{\text{cm, }}\,c = 25.8{\text{cm}}$$

Work Step by Step

$$\eqalign{ & B = {38^ \circ }40',\,\,\,a = 19.7{\text{cm,}}\,\,\,C = {91^ \circ }40' \cr & {\text{Find }}A \cr & A = {180^ \circ } - B - C \cr & A = {180^ \circ } - {38^ \circ }40' - {91^ \circ }40' \cr & A = {49^ \circ }40' \cr & \cr & {\text{Use the law of sines to find side }}b \cr & \frac{a}{{\sin A}} = \frac{b}{{\sin B}} \cr & b = \frac{{a\sin B}}{{\sin A}} \cr & b = \frac{{19.7\sin \left( {{{38}^ \circ }40'} \right)}}{{\sin \left( {{{49}^ \circ }40'} \right)}} \cr & {\text{Use a calculator}} \cr & b = 16.1{\text{cm}} \cr & \cr & {\text{Use the law of sines to find side }}c \cr & \frac{a}{{\sin A}} = \frac{c}{{\sin C}} \cr & c = \frac{{a\sin C}}{{\sin A}} \cr & c = \frac{{19.7\sin \left( {{{91}^ \circ }40'} \right)}}{{\sin \left( {{{49}^ \circ }40'} \right)}} \cr & c = 25.8{\text{cm}} \cr & \cr & {\text{Answer}} \cr & A = {49^ \circ }40',\,\,\,b = 16.1{\text{cm, }}\,c = 25.8{\text{cm}} \cr} $$
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