Chapter 8 - Application of Trigonometry - 8.1 The Law of Sines - 8.1 Exercises - Page 755: 45

$$\eqalign{ & {B_1} = 49^\circ 20',\,\,\,\,{C_1} = {92^ \circ }00',\,\,\,\,{c_1} = 15.5{\text{m}},\,\,\, \cr & {B_2} = 130^\circ 40',\,\,{C_2} = 10^\circ 40',\,\,{c_2} = 2.88{\text{m}},\,\,\, \cr} $$

$$\eqalign{ & A = {\text{38}}^\circ {\text{4}}0\prime ,\,\,\,a = {\text{9}}.{\text{72 m}},\,\,\,b = {\text{11}}.{\text{8 m}} \cr & {\text{Use the law of sines to find the angle of }}B \cr & \frac{{\sin B}}{b} = \frac{{\sin A}}{a} \cr & \sin B = \frac{{b\sin A}}{a} \cr & \sin B = \frac{{11.8\sin \left( {{\text{38}}^\circ {\text{4}}0\prime } \right)}}{{{\text{9}}.{\text{72}}}} \cr & {\text{Use a calculator}} \cr & \sin B = 0.7584881144 \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 & {B_1} = {\sin ^{ - 1}}\left( {0.7584881144} \right) \cr & {\text{Use the inverse sine function}} \cr & {B_1} = 49^\circ 20' \cr & \cr & {\text{Supplementary angles have the same sine value}},{\text{ so another }} \cr & {\text{possible value of angle }}B{\text{ is}} \cr & {B_2} = {180^ \circ } - 49^\circ 20' \cr & {B_2} = 130^\circ 40' \cr & \cr & {\text{Calculating }}{C_1} \cr & {C_1} = {180^ \circ } - {B_1} - A \cr & {C_1} = {180^ \circ } - 49^\circ 20' - {\text{38}}^\circ {\text{4}}0\prime \cr & {C_1} = 92^\circ 00' \cr & \cr & {\text{Calculating }}{C_2} \cr & {C_2} = {180^ \circ } - {B_2} - A \cr & {C_2} = {180^ \circ } - 130^\circ 40' - {\text{38}}^\circ {\text{4}}0\prime \cr & {C_2} = 10^\circ 40' \cr & \cr & {\text{Calculate }}{c_1}{\text{ using the law of sines}} \cr & \frac{{{c_1}}}{{\sin {C_1}}} = \frac{a}{{\sin A}} \cr & {c_1} = \frac{{a\sin {C_1}}}{{\sin A}} \cr & {c_1} = \frac{{{\text{9}}.{\text{72}}\sin \left( {92^\circ 00'} \right)}}{{\sin \left( {{\text{38}}^\circ {\text{4}}0\prime } \right)}} \cr & {c_1} \approx 15.5{\text{m}} \cr & \cr & {\text{Calculate }}{c_2}{\text{ using the law of sines}} \cr & \frac{{{c_2}}}{{\sin {C_2}}} = \frac{a}{{\sin A}} \cr & {c_2} = \frac{{a\sin {C_2}}}{{\sin A}} \cr & {c_2} = \frac{{{\text{9}}.{\text{72}}\sin \left( {10^\circ 40'} \right)}}{{\sin \left( {{\text{38}}^\circ {\text{4}}0\prime } \right)}} \cr & {c_2} \approx 2.88{\text{m}} \cr & \cr & {\text{Answer}} \cr & {B_1} = 49^\circ 20',\,\,\,\,{C_1} = {92^ \circ }00',\,\,\,\,{c_1} = 15.5{\text{m}},\,\,\, \cr & {B_2} = 130^\circ 40',\,\,{C_2} = 10^\circ 40',\,\,{c_2} = 2.88{\text{m}},\,\, \cr} $$