2. To find the distance AB across a river, a distance BC of 319 m is laid off on one side of the river. It is found that B = 104.6° and C = 14.4°. Find AB.

In a diagram,
Notice that we can calculate angle A by using the fact that the sum of the inner angles of a triangle is 180°; therefore, in our case,
[tex]\begin{gathered} \angle A=180-\angle B-\angle C=180-104.6-14.4=61 \\ \Rightarrow\angle A=61\degree \end{gathered}[/tex]On the other hand, the law of sines states that
[tex]\begin{gathered} \frac{sinX}{x}=\frac{sinY}{y}=\frac{sinZ}{z} \\ x,y,z\rightarrow sides\text{ of a triangle} \\ X\rightarrow\text{ opposite angle to side x} \\ Similarly\text{ for Y and Z} \end{gathered}[/tex]Thus, applying the law of sines to the triangle above,
[tex]\begin{gathered} \frac{sinA}{BC}=\frac{sinC}{AB} \\ \Rightarrow AB=BC\frac{sinC}{sinA} \end{gathered}[/tex]Hence,
[tex]\begin{gathered} \Rightarrow AB=319*(\frac{sin(14.4)}{sin(61)}) \\ \Rightarrow AB=90.7 \end{gathered}[/tex]