How to apply the law of sines in a non right triangle?

Question

From http://www.sparknotes.com/testprep/books/sat2/math2c/chapter9section9.rhtml, I saw that you can apply the law of sines to solve the measures of all the variable values of a non-right triangle when you know 2 angles and the length of a side. However, the website only gave an example of application when you know 2 sides and a length of an angle. In this case, because I know 2 angles, and they are both real numbers which do not contain variables, I can find the 3rd angle. When given this information, how could I apply the law of sines as seen on http://www.sparknotes.com/testprep/books/sat2/math2c/chapter9section9.rhtml?

Answer 1

Let's use the common convention that $\alpha$, $\beta$ and $\gamma$ denote the measures of the angles at the vertices $A$, $B$ and $C$, respectively, whereas $a$, $b$ and $c$ denote the measures of the sides opposite to $A$, $B$ and $C$, respectively. Then the law of sines is $$ \frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma} $$ Suppose we know $\alpha$, $\beta$ and $c$ (knowing two angles means knowing all three of them, so we can use the ones adjacent to $c$).

Then $$ \frac{a}{\sin\alpha}=\frac{c}{\sin(\pi-\alpha-\beta)} $$ or $$ a=\frac{c\sin\alpha}{\sin(\alpha+\beta)} $$ and, similarly, $$ b=\frac{c\sin\beta}{\sin(\alpha+\beta)}. $$

Answer 2

Law of Sines says that $\frac{\sin \alpha}{a}$ is always the same for any angle $\alpha$ and the length of the side across from it, $a$. You said you know all three angles and one side. Call the side you know $b$ and let $\beta$ be the angle across from it. Then for any other side $c$ and the angle $\gamma$ across from it (which you also know), you know that $$ \frac{\sin \beta}{b}=\frac{\sin \gamma}{c}. $$ Solve for $c$.