Chapter 8 - Application of Trigonometry - 8.1 The Law of Sines - 8.1 Exercises - Page 754: 33

$${B_1} = {49.1^ \circ },\,\,\,\,\,\,{B_2} = {130.9^ \circ },\,\,\,\,\,\,\,{C_1} = {101.2^ \circ },\,\,\,\,{C_2} = {19.4^ \circ }$$

$$\eqalign{ & A = {29.7^ \circ },\,\,\,b = 41.5{\text{ft,}}\,\,\,a = 27.2{\text{ft}} \cr & \cr & {\text{Use the law of sines to find the angle of }}B \cr & \frac{a}{{\sin A}} = \frac{b}{{\sin B}} \cr & \sin B = \frac{{b\sin A}}{a} \cr & \sin B = \frac{{41.5\sin \left( {{{29.7}^ \circ }} \right)}}{{27.2}} \cr & {\text{Use a calculator}} \cr & \sin B = 0.755938 \cr & \cr & {\text{There are two }}\,{\text{angles }}B\,{\text{between }}\,{{\text{0}}^ \circ }{\text{ and 18}}{{\text{0}}^ \circ }{\text{ that satisfy this }} \cr & {\text{condition}}{\text{.}} \cr & {B_1} = {\sin ^{ - 1}}\left( {0.755938} \right) \cr & {\text{Use the inverse sine function}} \cr & {B_1} = {49.1^ \circ } \cr & \cr & {\text{Supplementary angles have the same sine value}},{\text{ so another }} \cr & {\text{possible value of angle }}B{\text{ is}} \cr & {B_2} = {180^ \circ } - {49.1^ \circ } \cr & {B_2} = {130.9^ \circ } \cr & \cr & {\text{Calculating }}{C_1} \cr & {C_1} = {180^ \circ } - A - {B_1} \cr & {C_1} = {180^ \circ } - {29.7^ \circ } - {49.1^ \circ } \cr & {C_1} = {101.2^ \circ } \cr & \cr & {\text{Calculating }}{C_2} \cr & {C_2} = {180^ \circ } - A - {B_2} \cr & {C_2} = {180^ \circ } - {29.7^ \circ } - {130.9^ \circ } \cr & {C_2} = {19.4^ \circ } \cr & \cr & {\text{Answer}} \cr & {B_1} = {49.1^ \circ },\,\,\,\,\,\,{B_2} = {130.9^ \circ },\,\,\,\,\,\,\,{C_1} = {101.2^ \circ },\,\,\,\,{C_2} = 19.4 \cr} $$