Find Side c of a Triangle: Formula, Steps & Examples

When You Cannot Reach Across the Gap

A surveyor at the edge of a canyon cannot stretch a tape measure across to the far wall. Yet the distance across gets reported to the centimetre. The trick is old and quietly brilliant: measure the two sides you can reach and the angle between them, then compute the third side you cannot. That third side is what mapmakers, radar operators, and roof carpenters spend their day finding, and the same short calculation gives it every time.

What Does "Finding Side c" Actually Mean?

Finding side c means computing the length of the third side of a triangle when you already know enough of the other measurements. Two tools cover almost every case. If the triangle has a right angle, the Pythagoras theorem gives $c^2 = a^2 + b^2$. If there is no right angle but you know two sides and the angle between them, the Law of Cosines gives $c^2 = a^2 + b^2 - 2ab\cos C$. Everything else is choosing the right tool and labelling carefully.

Notice the naming rule that makes both formulas work: side $c$ sits opposite angle $C$. Keep that pairing and the algebra takes care of itself. Break it and you will solve for the wrong side without any warning.

How Do You Know Which Formula to Use?

This is the single most-asked version of the question on homework forums, so answer it before touching numbers.

• Right angle present, two legs known, use Pythagoras. The hypotenuse is always the side opposite the right angle.

• Two sides and the included angle known (SAS), no right angle, use the Law of Cosines. It is the general tool; Pythagoras is just its special case when $\cos C = \cos 90^\circ = 0$.

• Only two sides and nothing else, you cannot pin down a single value for $c$. More on that below, because it is a genuine trap.

How Do You Find Side c Step by Step?

The procedure is short. For the Law of Cosines, which handles every case:

1. Label the two known sides $a$ and $b$, and the angle between them as $\angle C$.

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

3. Evaluate the right-hand side, keeping the sign of $\cos C$ intact.

4. Take the positive square root: $c = \sqrt{a^2 + b^2 - 2ab\cos C}$.

For a right triangle, drop the cosine term entirely and work with $c^2 = a^2 + b^2$.

Examples of Finding Side c of a Triangle

Each example builds on the last. The numbers stay clean early and turn realistic by the end.

Example 1

In a right triangle, the two legs are $a = 3$ and $b = 4$. Find the hypotenuse $c$.

Because the angle between the legs is $90^\circ$, use Pythagoras.

$$c^2 = a^2 + b^2 = 3^2 + 4^2$$

$$c^2 = 9 + 16 = 25$$

$$c = \sqrt{25} = 5$$

Final answer: $c = 5$.

Example 2

A triangle has $a = 7$, $b = 10$, and the included angle $\angle C = 60^\circ$. Find side $c$.

The tempting move is to treat this like Example 1 and write $c^2 = 7^2 + 10^2 = 149$, giving $c \approx 12.2$. Take a moment with that. The $60^\circ$ angle is sharper than a right angle, which should pull the far corner inward and make $c$ shorter than the right-angle value, yet $12.2$ is longer than either side. Pythagoras assumed a right angle that is not there.

Now the rescue. Use the Law of Cosines with $\cos 60^\circ = 0.5$.

$$c^2 = 7^2 + 10^2 - 2(7)(10)\cos 60^\circ$$

$$c^2 = 49 + 100 - 140(0.5)$$

$$c^2 = 149 - 70 = 79$$

$$c = \sqrt{79} \approx 8.89$$

Final answer: $c \approx 8.89$, shorter than the flawed $12.2$, exactly as the geometry predicted.

Example 3

A right triangle has hypotenuse $13$ and one leg $5$. Find the other leg $c$.

Here $c$ is a leg, so rearrange Pythagoras.

$$c^2 = 13^2 - 5^2$$

$$c^2 = 169 - 25 = 144$$

$$c = \sqrt{144} = 12$$

Final answer: $c = 12$. (You may recognise $5, 12, 13$ as a Pythagorean triple.)

Example 4

A triangle has $a = 5$, $b = 8$, and an obtuse included angle $\angle C = 120^\circ$. Find side $c$.

Obtuse angles are where signs matter. $\cos 120^\circ = -0.5$, a negative value.

$$c^2 = 5^2 + 8^2 - 2(5)(8)\cos 120^\circ$$

$$c^2 = 25 + 64 - 80(-0.5)$$

$$c^2 = 89 + 40 = 129$$

$$c = \sqrt{129} \approx 11.36$$

Final answer: $c \approx 11.36$. The subtraction became an addition because $\cos C$ was negative, so $c$ came out longer than both sides, correct for a wide, opened-up angle.

Example 5

Can sides of length $3$ and $4$ produce a third side $c = 9$?

No calculation of $c$ is possible here, because no angle is given. The question is really about whether such a triangle can exist at all, and the triangle inequality theorem settles it: the two shorter sides must sum to more than the longest.

$$3 + 4 = 7 < 9$$

Final answer: No triangle exists. With only two sides and no angle, $c$ is not a single number, it can range anywhere from just above $|4 - 3| = 1$ to just under $4 + 3 = 7$. You need the angle (or a right-angle guarantee) to fix one value.

Example 6

A surveyor stands at a point and measures $120$ m to one landmark and $150$ m to another. The angle between the two sightlines is $70^\circ$. How far apart are the landmarks?

The distance between the landmarks is the unknown side $c$, opposite the $70^\circ$ angle at the surveyor. Use the Law of Cosines with $\cos 70^\circ \approx 0.342$.

$$c^2 = 120^2 + 150^2 - 2(120)(150)\cos 70^\circ$$

$$c^2 = 14400 + 22500 - 36000(0.342)$$

$$c^2 = 36900 - 12312 = 24588$$

$$c = \sqrt{24588} \approx 156.8$$

Final answer: the landmarks are about $156.8$ m apart, a distance nobody had to walk.

Why Does the Law of Cosines Solve the Unreachable Distance?

The Law of Cosines earns its keep wherever a direct measurement is impossible: across water, across a battlefield, across the vacuum between a radar dish and an aircraft. Its real origin is triangulation, the backbone of mapmaking. From two known positions and the angle to a distant target, surveyors chained triangle after triangle across whole continents, computing each third side rather than pacing it out.

• Surveying: the Law of Cosines (and its partner, the law of sines) let a chain of triangles cover terrain no tape could span.

• Navigation and radar: the distance between two moving objects follows from two ranges and the angle between them.

• The right-angle special case: when $\angle C = 90^\circ$, the cosine term vanishes and you are left with plain Pythagoras, the general law contains the familiar one.

For the formal statement and derivation, the Wolfram MathWorld entry on the Law of Cosines is a solid authoritative reference. The cosine itself comes from the trigonometric ratios of an angle, so a shaky grasp of $\cos\theta$ is worth shoring up first. When the angle happens to be $30^\circ$, $60^\circ$, or $90^\circ$, you can often skip the calculator entirely and read the ratios off a 30-60-90 triangle.

What Are the Most Common Mistakes When Finding Side c?

Mistake 1: Using Pythagoras on a triangle with no right angle

Where it slips in: any SAS problem where the given angle is not $90^\circ$, especially under time pressure.

Don't do this: write $c^2 = a^2 + b^2$ whenever you see two sides. That silently assumes a right angle that the problem never promised.

The correct way: check the angle first. Only a genuine $90^\circ$ angle licenses Pythagoras; everything else needs the $-2ab\cos C$ term.

Mistake 2: Pairing the wrong angle with side c

Where it slips in: triangles labelled out of order, or diagrams where the unknown side is not opposite the given angle.

Don't do this: plug in whichever angle is written first. If the angle you use is not opposite the side you want, the formula solves for a different side entirely.

The correct way: confirm that $\angle C$ sits between the two known sides and opposite side $c$. The step of matching angle to opposite side feels skippable, and it is exactly where the answer goes wrong.

Mistake 3: Dropping the sign of the cosine for obtuse angles

Where it slips in: whenever $\angle C$ is larger than $90^\circ$, so $\cos C$ is negative.

Don't do this: treat $-2ab\cos C$ as a subtraction out of habit. When $\cos C < 0$, that whole term becomes a positive addition, and forcing a subtraction shrinks $c$ when it should grow.

The correct way: carry the sign of $\cos C$ through untouched. A wide angle opens the triangle up, so $c$ must come out longer than either side.

Conclusion

• Find side c with Pythagoras ($c^2 = a^2 + b^2$) for right triangles and the Law of Cosines ($c^2 = a^2 + b^2 - 2ab\cos C$) for any triangle with two sides and the included angle.

• Keep side $c$ opposite $\angle C$, and carry the sign of $\cos C$ (negative for obtuse angles) all the way through.

• With only two sides and no angle, side $c$ has a range set by the triangle inequality, not a single value.

• The same calculation that solves a textbook triangle is what lets surveyors measure across gaps they cannot cross.

To work through these methods with a teacher, explore Bhanzu's geometry tutor or high school math tutor sessions, or browse the live math classes online.

A Practical Next Step

Practice these problems to solidify your understanding: rework Examples 2 and 4 changing only the angle (try $45^\circ$, then $135^\circ$) and watch how side $c$ responds. If you get stuck on the sign of the cosine, return to Mistake 3 above. Want a live trainer to check your labelling and method choice in real time? Book a free demo class.

Read More

• Pythagorean triples, the whole-number side sets that make right-triangle problems instant.

• Converse of the Pythagoras theorem, how to test whether a triangle is right-angled from its sides alone.

• Right triangle formulas, a compact reference for sides, area, and height.

• Isosceles right triangle, the 45-45-90 case and its fixed side ratio.

• Types of triangles, classifying triangles by their sides and angles.