# The Law of Cosines

Using the Sine and Cosine Rule To Solve Unknown Angles

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• 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: ${b}^{2} = 84 \mathmr{and} b = \sqrt{84} = 9 , 2 c m$

• 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.

• Find C using the Law of Cosines.

${c}^{2} = {b}^{2} + {a}^{2} - 2 a b C o s \left(C\right)$
This can also be written as:
${a}^{2} = {b}^{2} + {c}^{2} - 2 b c C o s A$
${b}^{2} = {a}^{2} + {c}^{2} - 2 a c C o s B$

Say you have a triangle with Angles A, B, and C and sides a, b, and c as shown below.

In order to find the angle, the problem will most likely give you the length of the 3 sides.

Just for explanation purposes, let's give these side lengths values.
a=9, c=8, and b=5.

If we substituted all of these values into our Law of Cosines equation, the new equation would look like:
${8}^{2} = {5}^{2} + {9}^{2} - 2 \left(5\right) \left(9\right) C o s \left(C\right)$

You wold then simplify the equation so that it comes down to
-42=-90Cos(C)

Isolate the Cos(C):
Cos(C)=42/90

You now have to take the inverse of Cosine to isolate Angle C.
$C = C o {s}^{-} 1 \left(\frac{42}{90}\right)$
Round to however your teacher wants you to, but you should get about 62.18 degrees.

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