# Solving Triangles (finding missing sides/angles given 3 sides/angles)

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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(intended to be used as a generalization for closing specific-case triangle-solving questions) – Isaac Feb 7 '12 at 3:32
Brilliant idea! BTW, is it true that such questions are added to a list at Meta and a (faq) tag added to them? – user21436 Feb 7 '12 at 3:36
+1 for the motive! – user21436 Feb 7 '12 at 3:37
@KannappanSampath: Yes: meta.math.stackexchange.com/questions/1868/… (I personally prefer to wait and see how the question does and what kind of other answer(s) and comments it may attract before adding it to that list and adding the faq tag.) – Isaac Feb 7 '12 at 3:39

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.

## SSS

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:

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

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

$$C=180°-A-B$$

## SAS

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:

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

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

$$B=180°-A-C$$

## ASA or AAS

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:

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

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

## SSA

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:

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

## AAA

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.

## Reference

### Law of Cosines

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

### Law of Sines

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

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Good, it will be useful. Maybe it would be helpful to state the Law of Cosines and the Law of Sines in standard form. And perhaps add links to proofs for the curious. – André Nicolas Feb 7 '12 at 3:42
@AndréNicolas: I think you're right. Does the added "Reference" section at the end satisfy your suggestion? – Isaac Feb 7 '12 at 3:46
Sure, it is fine. Just thought it would be nice to have something that pointed beyond how-to. – André Nicolas Feb 7 '12 at 4:48
In SSS and SAS, for the 2nd step I'd use Law of Sines instead of Law of Cosines. It's somewhat less computation. – Gerry Myerson Feb 7 '12 at 4:55
@GerryMyerson: While it's less computation, there's substantial risk (especially for students) in missing what should be an obtuse angle when using the Law of Sines; by using the Law of Cosines, we can avoid having to consider whether we should be using $\arcsin(\cdots)$ or $\pi-\arcsin(\cdots)$. – Isaac Feb 7 '12 at 4:57

## protected by Nate EldredgeNov 17 '13 at 17:55

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