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What is a general procedure for “solving” a triangle—that is, for finding the unknown side lengths and angle measures given three side lengths and/or angle measures?

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First, some notation: let $A$, $B$, and $C$ be the measures of the three angles and let $a$, $b$, and $c$ be the lengths of the sides opposite those angles, respectively. Now, let’s look case-by-case at the possible sets of information we could have.

Let’s start with the case where we know the three sides, $a$, $b$, and $c$. We can use the Law of Cosines, in the form where it is solved for the cosine of an angle, to find the measures of two of the angles, then use the fact that the sum of the measures of the angles is 180° to find the third:

$$A=\arccos\left(\frac{b^2+c^2-a^2}{2bc}\right)$$

$$B=\arccos\left(\frac{a^2+c^2-b^2}{2ac}\right)$$

$$C=180°-A-B$$

If we know two sides and the measure of the angle they include, say $a$, $b$, and $C$, we can use the Law of Cosines to find the unknown side length, then pick up with the SSS process to find the unknown angles:

$$c=\sqrt{a^2+b^2-2ab\cos C}$$

$$A=\arccos\left(\frac{b^2+c^2-a^2}{2bc}\right)$$

$$B=180°-A-C$$

If we know the measures of two angles, we can find the measure of the third angle using $A+B+C=180°$, so let’s assume we know all three angle measures $A$, $B$, and $C$, and the side length $a$. We can use the Law of Sines to find each of the unknown side lengths:

$$b=\frac{a\sin B}{\sin A}$$

$$c=\frac{a\sin C}{\sin A}$$

If we know two side lengths and the measure of an angle that isn’t included between the two known sides, say $a$, $b$, and $A$, we can start by using the Law of Sines to find $B$, but this may give two solutions:

$$\sin B=\frac{b\sin A}{a}$$

If $\sin B=\frac{b\sin A}{a}>1$, then there is no solution and the given information does not determine a triangle (it is impossible for the given information to describe a triangle).

If $\sin B=\frac{b\sin A}{a}=1$, then $B$ is a right angle and the given information determines a single triangle.

If $\sin B=\frac{b\sin A}{a}<1$, then there are two solutions for $B$:

$$B_1=\arcsin\left(\frac{b\sin A}{a}\right)\text{ or }B_2=180°-B_1$$

In each case, we can use $A+B+C=180°$ to determine $C$:

$$C_1=180°-A-B_1\text{ or }C_2=180°-A-B_2$$

At this point, $B_1$ and $C_1$ will definitely describe a triangle, but if $C_2\le0$ then $B_2$ and $C_2$ do not describe a triangle. Whether we only have the $B_1$ case or both the $B_1$ and $B_2$ cases, we can use the Law of Cosines to find the unknown side:

$$c=\sqrt{a^2+b^2-2ab\cos C}$$

Knowing three angles only determines the triangle up to similarity—that is, if we don’t know at least one length, we’re not going to be able to find any lengths.

Wikipedia MathWorld

$$c^2=a^2+b^2-2ab\cos C$$

$$\cos C=\frac{a^2+b^2-c^2}{2ab}$$

Wikipedia MathWorld

$$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$$

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