Answer
$$a = 156{\text{cm}},{\text{ }}B = {64^ \circ }50',\,\,C = {34^ \circ }30'$$
Work Step by Step
$$\eqalign{
& A = {80^ \circ }40'{\text{,}}\,\,\,b = 143{\text{cm,}}\,\,\,c = 89.6{\text{cm}} \cr
& {\text{Use the Law of cosines to find }}a \cr
& {a^2} = {b^2} + {c^2} - 2bc\cos A \cr
& {\text{Substitute}} \cr
& {a^2} = {\left( {143} \right)^2} + {\left( {89.6} \right)^2} - 2\left( {143} \right)\left( {89.6} \right)\cos \left( {{{80}^ \circ }40'} \right) \cr
& {\text{Use a calculator}} \cr
& {a^2} \approx 24321.25341 \cr
& {\text{Take square roots and choose the positive root}} \cr
& a \approx 156{\text{cm}} \cr
& \cr
& {\text{Calculate the angle }}B{\text{ using the law of sines}} \cr
& \frac{b}{{\sin B}} = \frac{a}{{\sin A}} \cr
& \sin B = \frac{{b\sin A}}{a} \cr
& \sin B = \frac{{143\sin \left( {{{80}^ \circ }40'} \right)}}{{156}} \cr
& {\text{Use a calculator}} \cr
& \sin B \approx 0.9045314044 \cr
& {\text{Use the inverse sine function}} \cr
& B = {64^ \circ }50' \cr
& \cr
& {\text{Calculate }}C \cr
& C = {180^ \circ } - A - B \cr
& C = {180^ \circ } - {80^ \circ }40' - {64^ \circ }50' \cr
& C = {34^ \circ }30' \cr
& \cr
& {\text{Answer}} \cr
& a = 156{\text{cm}},{\text{ }}B = {64^ \circ }50',\,\,C = {34^ \circ }30' \cr} $$
