Law of Cosines

Revision as of 15:08, 7 October 2007 by Temperal (talk | contribs) (Right Triangle: right triange)
This is an AoPSWiki Word of the Week for Oct 4-Oct 10

The Law of Cosines is a theorem which relates the side-lengths and angles of a triangle. For a triangle with edges of length $a$, $b$ and $c$ opposite angles of measure $A$, $B$ and $C$, respectively, the Law of Cosines states:

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

$b^2 = a^2 + c^2 - 2ac\cos B$

$a^2 = b^2 + c^2 - 2bc\cos A$

In the case that one of the angles has measure $90^\circ$ (is a right angle), the corresponding statement reduces to the Pythagorean Theorem.

Proofs

Acute Triangle


An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.



Info: a, b, and c are the side lengths, and C is the angle measure opposite side C. f is the height from angle C to side c, and d and e are the lengths that c is split into by f.

We use the pythagorean theorem:

\[a^2+b^2-2f^2=d^2+e^2\]

We are trying to get $a^2+b^2-2f^2+2de$ on the LHS, because then the RHS would be $c^2$.

We use the addition rule for cosines and get:

\[\cos{C}=\dfrac{f}{a}*\dfrac{f}{b}-\dfrac{d}{a}*\dfrac{e}{b}=\dfrac{f^2-de}{ab}\]

We multiply by -2ab and get:

\[2de-2f^2=-2ab\cos{C}\]

Now remember our equation?

\[a^2+b^2-2f^2+2de=c^2\]

We replace the $-2f^2+2de$ by $-2ab\cos{C}$ and get:

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

We can use the same argument on the other sides.

Right Triangle

Since $C=90^{circ}$, $\cos C=0$, so the expression reduces to the Pythagorean Theorem. You can find several proofs of the Pythagorean Theorem here

Obtuse Triangle

See also