The Cosine Law

The sine law needs a side and its opposite angle. When you don’t have such a pair, as when you know two sides and the angle between them, or all three sides, the cosine law steps in. It’s the Pythagorean theorem, upgraded to work in any triangle. All angles are in degrees.

Key ideas

The cosine law

In any triangle $ABC$ (labelled as below):

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

The same pattern works for every side: $a^2 = b^2 + c^2 - 2bc\cos A$ and $b^2 = a^2 + c^2 - 2ac\cos B$. The side on the left is opposite the angle in the cosine.

Finding an angle

Rearranging to find an angle from three sides:

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

When to use it

You know	Use
two sides and the angle between them (SAS)	cosine law, to find the third side
all three sides (SSS)	cosine law, to find an angle
a side and its opposite angle, plus one more	sine law

Connection to Pythagoras

If $\angle C = 90^\circ$, then $\cos C = 0$ and the law becomes $c^2 = a^2 + b^2$. The $-2ab\cos C$ term corrects for angles that aren’t right angles.

The cosine law has no ambiguous case: $\cos^{-1}$ gives values from $0^\circ$ to $180^\circ$, and a negative cosine correctly gives an obtuse angle.

Worked examples

Example 1: Two sides and the included angle (SAS)

In $\triangle ABC$, $a = 8$, $b = 11$, and $\angle C = 52^\circ$. Find $c$ to two decimal places.

Solution.

$$\begin{aligned} c^2 &= 8^2 + 11^2 - 2(8)(11)\cos 52^\circ \\ &= 185 - 176\cos 52^\circ \\ &\approx 76.64 \end{aligned}$$

So $c \approx \sqrt{76.64} \approx 8.75$.

Example 2: Three sides (SSS)

A triangle has sides $a = 7$, $b = 9$, and $c = 12$. Find its largest angle.

Solution. The largest angle is opposite the longest side, so find $\angle C$:

$$\cos C = \frac{7^2 + 9^2 - 12^2}{2(7)(9)} = \frac{49 + 81 - 144}{126} = \frac{-14}{126} \approx -0.1111$$

$\angle C \approx \cos^{-1}(-0.1111) \approx 96.4^\circ$. The negative cosine tells you the angle is obtuse.

Example 3: Back to Pythagoras

Use the cosine law to find $c$ when $a = 3$, $b = 4$, and $\angle C = 90^\circ$.

Solution.

$$c^2 = 9 + 16 - 2(3)(4)\cos 90^\circ = 25 - 0 = 25 \quad\Rightarrow\quad c = 5$$

That’s the familiar $3$–$4$–$5$ right triangle.

Example 4: Two hikers

Two hikers leave camp on straight paths $65^\circ$ apart. One walks $4.5$ km and the other walks $6$ km. How far apart are they?

Solution. The angle between the two known sides is $65^\circ$:

$$d^2 = 4.5^2 + 6^2 - 2(4.5)(6)\cos 65^\circ = 56.25 - 54\cos 65^\circ \approx 33.43$$

$d \approx 5.78$ km.

Common mistakes

Using an angle that isn’t between the two sides. For $c^2 = a^2 + b^2 - 2ab\cos C$, angle $C$ must be the angle between sides $a$ and $b$ (opposite $c$).

Order of operations. $185 - 176\cos 52^\circ$ means multiply first, then subtract. It isn’t $(185 - 176)\cos 52^\circ$.

Forgetting the square root. The formula gives $c^2$. Take the square root at the end.

Picking the wrong angle to find first. When solving a whole triangle from three sides, find the largest angle first with the cosine law. Then the remaining angles are acute, and the sine law is safe to use.

Practice

1. (Warm-up) Which law would you use first?

• (a) two sides and the angle between them
• (b) two angles and a side
• (c) three sides
• (d) two sides and an angle opposite one of them

Solution

(a) Cosine law. (b) Sine law. (c) Cosine law. (d) Sine law (watch for the ambiguous case).

2. (Warm-up) Find $c$ if $a = 3$, $b = 5$, and $\angle C = 60^\circ$. Give an exact answer and a decimal.

Solution

$$c^2 = 9 + 25 - 2(3)(5)\left(\tfrac{1}{2}\right) = 34 - 15 = 19$$

$c = \sqrt{19} \approx 4.36$.

3. (Warm-up) A triangle has all three sides equal to $5$. Use the cosine law to find one of its angles.

Solution

$$\cos C = \frac{25 + 25 - 25}{2(5)(5)} = \frac{25}{50} = \frac{1}{2}$$

So $\angle C = 60^\circ$, as expected for an equilateral triangle.

4. (Core) In $\triangle ABC$, $b = 14$, $c = 9$, and $\angle A = 110^\circ$. Find $a$.

Solution

$$a^2 = 14^2 + 9^2 - 2(14)(9)\cos 110^\circ = 277 - 252\cos 110^\circ \approx 363.19$$

$a \approx 19.06$. ($\cos 110^\circ$ is negative, so this side is longer than it would be in a right triangle.)

5. (Core) A triangle has sides $10$, $13$, and $15$. Find its smallest angle.

Solution

The smallest angle is opposite the shortest side, $10$:

$$\cos\theta = \frac{13^2 + 15^2 - 10^2}{2(13)(15)} = \frac{294}{390} \approx 0.7538$$

$\theta \approx 41.1^\circ$.

6. (Core) Find all three angles of a triangle with sides $6$, $8$, and $11$.

Solution

Largest angle first (opposite $11$):

$$\cos C = \frac{36 + 64 - 121}{2(6)(8)} = \frac{-21}{96} \quad\Rightarrow\quad \angle C \approx 102.6^\circ$$

Angle opposite $6$:

$$\cos A = \frac{64 + 121 - 36}{2(8)(11)} = \frac{149}{176} \quad\Rightarrow\quad \angle A \approx 32.2^\circ$$

The third angle is about $180^\circ - 102.6^\circ - 32.2^\circ = 45.2^\circ$.

7. (Core) A triangular garden has two sides of $12$ m and $15$ m with a $72^\circ$ angle between them. How much fencing is needed to go all the way around?

Solution

$$c^2 = 144 + 225 - 2(12)(15)\cos 72^\circ = 369 - 360\cos 72^\circ \approx 257.75$$

$c \approx 16.05$ m. The fencing needed is about $12 + 15 + 16.05 = 43.05$ m.

8. (Challenge) Explain what the cosine law says about $c^2$ compared with $a^2 + b^2$ when $\angle C$ is acute, right, and obtuse.

Solution

$c^2 = a^2 + b^2 - 2ab\cos C$, and $2ab$ is positive.

• Acute $C$: $\cos C \gt 0$, so $c^2 \lt a^2 + b^2$.
• Right $C$: $\cos C = 0$, so $c^2 = a^2 + b^2$ (Pythagoras).
• Obtuse $C$: $\cos C \lt 0$, so $c^2 \gt a^2 + b^2$.

The bigger the angle, the longer the opposite side.

9. (Challenge) A parallelogram has sides of $6$ cm and $10$ cm with an angle of $50^\circ$ between them. Find the lengths of both diagonals.

Solution

The parallelogram’s angles are $50^\circ$ and $130^\circ$. Each diagonal is the third side of a triangle with sides $6$ and $10$.

Short diagonal ($50^\circ$): $d^2 = 36 + 100 - 120\cos 50^\circ \approx 58.87$, so $d \approx 7.67$ cm.

Long diagonal ($130^\circ$): $d^2 = 36 + 100 - 120\cos 130^\circ \approx 213.13$, so $d \approx 14.60$ cm.