Chapter 8 - Application of Trigonometry - 8.1 The Law of Sines - 8.1 Exercises - Page 755: 49

$$\eqalign{ & {A_1} = 53.23^\circ ,\,\,\,\,{C_1} = 87.09^\circ ,\,\,\,\,{c_1} = 37.16{\text{m}} \cr & {A_2} = 126.77^\circ ,\,\,{C_2} = 13.55^\circ ,\,\,{c_2} = 8.719 \cr} $$

$$\eqalign{ & B = {\text{39}}.{\text{68}}^\circ ,\,\,a = {\text{29}}.{\text{81 m}},\,\,\,b = {\text{23}}.{\text{76 m}} \cr & {\text{Use the law of sines to find the angle of }}A \cr & \frac{{\sin A}}{a} = \frac{{\sin B}}{b} \cr & \sin A = \frac{{a\sin B}}{b} \cr & \sin A = \frac{{{\text{29}}.{\text{81}}\sin \left( {{\text{39}}.{\text{68}}^\circ } \right)}}{{{\text{23}}.{\text{76}}}} \cr & {\text{Use a calculator}} \cr & \sin A = 0.8010800242 \cr & \cr & {\text{There are two }}\,{\text{angles }}A\,\,{\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( {0.8010800242} \right) \cr & {\text{Use the inverse sine function}} \cr & {A_1} = 53.23^\circ \cr & \cr & {\text{Supplementary angles have the same sine value}},{\text{ so another }} \cr & {\text{possible value of angle }}B{\text{ is}} \cr & {A_2} = {180^ \circ } - 53.23^\circ \cr & {A_2} = 126.77^\circ \cr & \cr & {\text{Calculating }}{C_1} \cr & {C_1} = {180^ \circ } - {A_1} - B \cr & {C_1} = {180^ \circ } - 53.23^\circ - {\text{39}}.{\text{68}}^\circ \cr & {C_1} = 87.09^\circ \cr & \cr & {\text{Calculating }}{C_2} \cr & {C_2} = {180^ \circ } - {A_2} - B \cr & {C_2} = {180^ \circ } - 126.77^\circ - {\text{39}}.{\text{68}}^\circ \cr & {C_2} = 13.55^\circ \cr & \cr & {\text{Calculate }}{c_1}{\text{ using the law of sines}} \cr & \frac{{{c_1}}}{{\sin {C_1}}} = \frac{b}{{\sin B}} \cr & {c_1} = \frac{{b\sin {C_1}}}{{\sin B}} \cr & {c_1} = \frac{{{\text{23}}.{\text{76}}\sin \left( {87.09^\circ } \right)}}{{\sin \left( {{\text{39}}.{\text{68}}^\circ } \right)}} \cr & {c_1} \approx 37.16{\text{m}} \cr & \cr & {\text{Calculate }}{c_2}{\text{ using the law of sines}} \cr & \frac{{{c_2}}}{{\sin {C_2}}} = \frac{b}{{\sin B}} \cr & {c_2} = \frac{{b\sin {C_2}}}{{\sin B}} \cr & {c_2} = \frac{{{\text{23}}.{\text{76}}\sin \left( {13.55^\circ } \right)}}{{\sin \left( {{\text{39}}.{\text{68}}^\circ } \right)}} \cr & {c_2} \approx 8.719 \cr & \cr & {\text{Answer}} \cr & {A_1} = 53.23^\circ ,\,\,\,\,{C_1} = 87.09^\circ ,\,\,\,\,{c_1} = 37.16{\text{m}} \cr & {A_2} = 126.77^\circ ,\,\,{C_2} = 13.55^\circ ,\,\,{c_2} = 8.719 \cr} $$