Since we have three points: A, B and C we can locate them in the vertexes of a triangle.
Now, from the observations made by the surveyor we know that the angle Notice that we also name the angles and distances. To find c, that is the distance between A and B, we can use the Law of sines
[tex]\frac{\sin\alpha}{a}=\frac{\sin\beta}{b}=\frac{\sin \gamma}{c}[/tex]but first we have to find the angle alpha. From the first two terms of the law we know that
[tex]\frac{\sin\alpha}{a}=\frac{\sin \beta}{b}[/tex]Plugging the values we have
[tex]\frac{\sin\alpha}{313\text{ ft}}=\frac{\sin 48.4}{527\text{ ft}}[/tex]Solving for alpha, we have
[tex]\begin{gathered} \frac{\sin\alpha}{313\text{ ft}}=\frac{\sin 48.4}{527\text{ ft}} \\ \sin \alpha=\frac{313\text{ ft}}{527\text{ ft}}\sin 48.4 \\ \alpha=\sin ^{-1}(\frac{313}{527}\sin 48.4) \\ \alpha=26.4^{\circ} \end{gathered}[/tex]Once we have the angle alpha, we can calculate the angle gamma. We know that the sum of all the interior angles of a triangle have to be 180, then
[tex]\alpha+\beta+\gamma=180[/tex]Plugging the values we know and solving for gamma we have
[tex]\begin{gathered} \alpha+\beta+\gamma=180 \\ 26.4+48.4+\gamma=180 \\ \gamma=180-26.4-48.4 \\ \gamma=105.2 \end{gathered}[/tex]Finally we can calculate the distance between the points A and B using the law of sines once again.
[tex]\frac{\sin\beta}{b}=\frac{\sin \gamma}{c}[/tex]Substituing the values we know and solving for c, we have
[tex]\begin{gathered} \frac{\sin48.4}{527\text{ ft}}=\frac{\sin105.2}{c} \\ c\frac{\sin48.4}{527\text{ ft}}=\sin 105.2 \\ c=\frac{\sin105.2}{\sin48.4}(527\text{ ft)} \\ c=680.1\text{ ft} \end{gathered}[/tex]Therefore the distance between the points A and B is 680.1 ft.
