Remember that the vertices and the sides of the triangles are usually labelled according to their locations as follows:
Side a is opposite to the angle A and so on.
According to the Law of Sines:
[tex]\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}[/tex]
Since A, a and b are known, then, we can use the Law of Sines to determie the value of B. First, solve for B using the inverse sine function:
[tex]\begin{gathered} \frac{a}{\sin(A)}=\frac{b}{\sin(B)} \\ \\ \Rightarrow\sin(B)=\frac{b}{a}\sin(A) \\ \\ \Rightarrow B=\sin^{-1}\left(\frac{b}{a}\sin(A)\right) \end{gathered}[/tex]
Replace b=21, a=34 and A=56º to find the value of B:
[tex]B=\sin^{-1}\left(\frac{21}{34}\sin(56º)\right)=30.8º[/tex]
Now, we know two internal angles of the triangle: A and B. Since the sum of the internal angles of any triangle must be equal to 180º, we can use that to find the valu of C:
[tex]\begin{gathered} A+B+C=180º \\ \Rightarrow C=180º-A-B \\ \Rightarrow C=180º-56º-30.8º \\ \therefore C=93.2º \end{gathered}[/tex]
Finally, sine we know A, a and C, we can use the law of sines again to find c:
[tex]\begin{gathered} \frac{c}{\sin(C)}=\frac{a}{\sin(A)} \\ \\ \Rightarrow c=\frac{\sin(C)}{\sin(A)}\times a=\frac{\sin(93.2º)}{\sin(56º)}\times34=40.947...\approx40.95 \end{gathered}[/tex]
Therefore, to two decimal places, the answers are:
[tex]\begin{gathered} B=30.80 \\ C=93.20 \\ c=40.95 \end{gathered}[/tex]
