We will use the law of sines and law of cosines shown below
[tex]\begin{gathered} c=\sqrt{a^2+b^2-2ab\cos C}\rightarrow\text{ law of cosines} \\ and \\ \frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}\rightarrow\text{ law of sines} \end{gathered}[/tex]
Therefore, in our case, finding side c.,
[tex]\begin{gathered} a=28,b=13,C=49\degree \\ \Rightarrow c=\sqrt{784+169-2*28*13cos(49\degree)} \\ \Rightarrow c\approx21.8 \end{gathered}[/tex]
Thus, side c is approximately 21.8km.
Finding the missing angles using the law of sines,
[tex]\begin{gathered} A=\cos^{-1}(\frac{c^2+b^2-a^2}{2bc}) \\ \Rightarrow A=\cos^{-1}(\frac{21.8^2+13^2-28^2}{2*13*21.8}) \\ \Rightarrow A\approx104.3 \\ \end{gathered}[/tex]
Similarly, in the case of angle B,
[tex]sinB=\frac{13}{21.8}sin(49\degree)\approx26.7\degree[/tex]
Therefore, the answers are
c=21.8km,
