Answer
$$A = {50^ \circ }50',\,\,\,\,B = {44^ \circ }40',\,\,\,C = {84^ \circ }30'$$
Work Step by Step
$$\eqalign{
& a = 965{\text{ft,}}\,\,\,b = 876{\text{ft,}}\,\,\,c = 1240{\text{ft}} \cr
& {\text{Use the law of cosines to solve for any angle of the triangle}} \cr
& {c^2} = {a^2} + {b^2} - 2ab\cos C \cr
& {\text{Solve for cos }}C \cr
& \cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} \cr
& {\text{Substitute}} \cr
& \cos C = \frac{{{{\left( {965} \right)}^2} + {{\left( {876} \right)}^2} - {{\left( {1240} \right)}^2}}}{{2\left( {965} \right)\left( {876} \right)}} \cr
& {\text{Use a calculator}} \cr
& \cos C \approx 0.095228547 \cr
& {\text{Use the inverse cosine function}} \cr
& C \approx {84^ \circ }30' \cr
& \cr
& {\text{Use the Law of sines to find the angle of }}B \cr
& \frac{{\sin B}}{b} = \frac{{\sin C}}{c} \cr
& \sin B = \frac{{b\sin C}}{c} \cr
& \sin B = \frac{{876\sin \left( {{{84}^ \circ }30'} \right)}}{{1240}} \cr
& {\text{Use a calculator}} \cr
& \sin B \approx 0.70331992498 \cr
& {\text{Use the inverse sine function}} \cr
& B \approx {44^ \circ }40' \cr
& \cr
& {\text{Calculate }}A \cr
& A = {180^ \circ } - B - C \cr
& A = {180^ \circ } - {44^ \circ }40' - {84^ \circ }30' \cr
& A = {50^ \circ }50' \cr
& \cr
& {\text{Answer}} \cr
& A = {50^ \circ }50',\,\,\,\,B = {44^ \circ }40',\,\,\,C = {84^ \circ }30' \cr} $$
