Answer
To solve the triangle here is to find the angle $B$, side $a$ and side $b$.
$$B=67.75^\circ, a=22.04mm, b=37.50mm$$
Work Step by Step
$$A=32^\circ57′,c=39.81mm,C=79^\circ18′ $$
$$B=?, a=?, b=?$$
1) Change the angle to complete degrees:
Remember that $60′=1^\circ$
Therefore,
$$A=32^\circ57′=32^\circ+(\frac{57}{60})^\circ=32^\circ+0.95^\circ=32.95^\circ$$
$$C=79^\circ18′=79^\circ+(\frac{18}{60})^\circ=79^\circ+0.3^\circ=79.3^\circ$$
2) Find $B$:
The sum of 3 angles in any triangle is $180^\circ$. That means,
$$32.95^\circ+B+79.3^\circ=180^\circ$$
$$112.25^\circ+B=180^\circ$$
$$B=67.75^\circ$$
3) Find $a$ and $c$
The law of sines for a triangle with side $a,b,c$ is
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$
First, to find $a$, we notice that the values of $c, C$ and $A$ are already known. So $a$ can be found using this
$$\frac{a}{\sin A}=\frac{c}{\sin C}$$
$$a=\frac{c\sin A}{\sin C}$$
$$a=\frac{39.81\sin32.95^\circ}{\sin79.3^\circ}$$
$$a=\frac{39.81\times0.544}{0.983}$$
$$a\approx22.04mm$$
To find $b$, we continue to notice that the values of $c, C$ and $B$ are already known. So $b$ can be found using this
$$b\sin B=c\sin C$$
$$b=\frac{c\sin B}{\sin C}$$
$$b=\frac{39.81\sin67.75^\circ}{\sin79.3^\circ}$$
$$b=\frac{39.81\times0.926}{0.983}$$
$$b\approx37.50mm$$
In conclusion:
$$B=67.75^\circ, a=22.04mm, b=37.50mm$$
