# Law of cosines

(Difference between revisions)
 Revision as of 11:04, 30 May 2011 (edit) (→Example Triangulation)← Previous diff Revision as of 11:21, 30 May 2011 (edit) (undo) (→Solution)Next diff → Line 68: Line 68: ---- ---- - Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure $B$, Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure $B$, Line 74: Line 73: $b^{2} = a^{2} + c^{2} - 2ab \cos B$ $b^{2} = a^{2} + c^{2} - 2ab \cos B$ + + Substituting in the appropriate side lengths gives us + + $(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$ + + Simplify for + + $36 (2) = 36 + 36 - 72 \cos B$ + + $72 = 72 - 72 \cos B$ + + Subtracting $72$ from both sides gives us + + $0 = - 72 \cos B$ + + Dividing both sides by $-72$ gives us + + $0 = \cos B$

## Revision as of 11:21, 30 May 2011

The law of cosines is a formula that helps in triangulation when two or three side lengths of a triangle are known. The formula relates all three side lengths of a triangle to the cosine of a particular angle.

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

When to use it: SAS, SSS.

## Proof

Let $\vartriangle ABC$ be oriented so that $C$ is at the origin, and $B$ is at the point$(a,0)$.

### Distance Formula

$distance = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$

$c$ is the distance from $A$ to $B$.

Substituting the appropriate points into the distance formula gives us

$c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}$

Squaring the inner terms, we have

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

Since $\cos^{2} C + \sin^{2} C = 1$,

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

Square both sides for

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

## Example Triangulation

Complete the triangle using the law of cosines.

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

### Solution

To find the side length $c$,

$c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ$

Simplify for

$c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ$

Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, substitution gives us

$c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}})$

Simplify for

$c^{2} =36 + 72 - 72$

$c^{2} =36$

Taking the square root of both sides gives us

$c =6$

Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure $B$,

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

Substituting in the appropriate side lengths gives us

$(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$

Simplify for

$36 (2) = 36 + 36 - 72 \cos B$

$72 = 72 - 72 \cos B$

Subtracting $72$ from both sides gives us

$0 = - 72 \cos B$

Dividing both sides by $-72$ gives us

$0 = \cos B$