# The Law of Cosines

## Key Questions

• The Law of Cosine can only be used if two sides and their enclosed angle are known.

In all other cases the Law of Sines has to be used.

• Yes, the Law of Cosines works for all triangles.

However, the proof depends on the shape a triangle, more precisely, how an altitude from some vertex falls onto the opposite side.

For example, consider a triangle $\Delta A B C$ with vertices $A$, $B$ and $C$, corresponding angles $\alpha$, $\beta$ and $\gamma$ and correspondingly opposite sides $a$, $b$ and $c$.
Let's prove the Law of Cosines that states:
${a}^{2} + {b}^{2} - 2 \cdot a \cdot b \cdot \cos \left(\gamma\right) = {c}^{2}$

Let's draw altitude $A H$ from vertex $A$ to an opposite side $B C$ with an intersection of this altitude and a side $B C$ at point $H$.
There are different cases of a location of point $H$ relatively to vertices $B$ and $C$.
It can lie in between vertices $B$ and $C$.
It can lie outside of $B C$ on a continuation of this side beyond vertex $B$ or beyond vertex $C$.

Assume that a base of this altitude, point $H$, is lying on the continuation of $B C$ beyond a point $C$ (so, $C$ is in between $B$ and $H$) and prove the Law of Cosines in this case. Other cases are similar to this one.

Let's use the following symbols for segments involved:
$A H$ is $h$
$B H$ is ${a}_{1}$
$C H$ is ${a}_{2}$
Then, since $C$ lies in between $B$ and $H$,
$a = {a}_{1} - {a}_{2}$ or ${a}_{1} = a + {a}_{2}$
Since both $\Delta A B H$ and $\Delta A C H$ are right triangles, by trigonometric dependency between hypotenuse, catheti and angles and by Pythagorean Theorem
$h = b \cdot \sin \left(\pi - \gamma\right) = b \cdot \sin \left(\gamma\right)$
${a}_{2} = b \cdot \cos \left(\pi - \gamma\right) = - b \cdot \cos \left(\gamma\right)$
${c}^{2} = {h}^{2} + {a}_{1}^{2} = {b}^{2} \cdot {\sin}^{2} \left(\gamma\right) + {\left(a - b \cdot \cos \left(y\right)\right)}^{2} =$
$= {b}^{2} \cdot {\sin}^{2} \left(\gamma\right) + {a}^{2} - 2 \cdot a \cdot b \cdot \cos \left(\gamma\right) + {b}^{2} \cdot {\cos}^{2} \left(\gamma\right) =$
$= {a}^{2} + {b}^{2} \left({\sin}^{2} \left(\gamma\right) + {\cos}^{2} \left(\gamma\right)\right) - 2 \cdot a \cdot b \cdot \cos \left(\gamma\right) =$
$= {a}^{2} + {b}^{2} - 2 \cdot a \cdot b \cdot \cos \left(\gamma\right)$

End of Proof

When point $H$ lies in between vertices $B$ and $C$ or on a continuation of side $B C$ beyond vertex B, the proof is similar.

See Unizor Trigonometry - Simple Identities - Law of Cosines for visual presentation and more detailed information.

$\cos \left(\gamma\right) = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b}$

#### Explanation:

We have
${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \left(\gamma\right)$
so
$2 a b \cos \left(\gamma\right) = {a}^{2} + {b}^{2} - {c}^{2}$
Isolating $\cos \left(\gamma\right)$
$\cos \left(\gamma\right) = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b}$

• Cosider the triangle:

(Picture source: Wikipedia)

you can relate the sides of this triangle in a kind of "extended" form of Pitagora's Theorem giving:

${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cdot \cos \left(\alpha\right)$
${b}^{2} = {a}^{2} + {c}^{2} - 2 a c \cdot \cos \left(\beta\right)$
${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cdot \cos \left(\gamma\right)$

As you can see you use this law when your triangle is not a right-angled one.

Example:
Consider the above triangle in which:
$a = 8 c m$
$c = 10 c m$
beta=60° therefore:
${b}^{2} = {a}^{2} + {c}^{2} - 2 a c \cdot \cos \left(\beta\right)$
b^2=8^2+10^2-2*8*10*cos(60°) but cos(60°)=1/2
so: ${b}^{2} = 84 \mathmr{and} b = \sqrt{84} = 9 , 2 c m$