Law of cosines

(Difference between revisions)
 Revision as of 16:13, 2 June 2011 (edit)← Previous diff Revision as of 10:53, 3 June 2011 (edit) (undo)Next diff → Line 10: Line 10: - 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 + The law of cosines is a formula that helps in solving triangles when two or three side lengths of a triangle are known. The formula combines the squares of two side lengths of a triangle and some offset, classified by the cosine of a particular angle, to calculate the square of the third side. For this reason, the law of cosines is often thought of as the generalization of the Pythagorean theorem, which only applies to right triangles. The law of cosines adds an extra term to the Pythagorean theorem so that a third side length of a triangle can be determined when there is no right angle. - particular angle. + - $c^{2} = a^{2} + b^{2} - 2ab \cos C$ + ::$c^{2} = a^{2} + b^{2} - 2ab \cos C$ When to use it: SAS, SSS. When to use it: SAS, SSS. Line 20: Line 19: ==Proof== ==Proof== + ===By the Pythagorean Theorem=== + An easy way to think of the law of cosines is as an extension of the Pythagorean theorem: + + ::$a^{2} + b^{2} = c^{2}$ + + [[Image:Pythagorean_cosines_proof.jpg]] + + + + ===Using the Distance Formula=== Let $\vartriangle ABC$ be oriented so that $C$ is at the origin, and $B$ is at the point$(a,0)$. Let $\vartriangle ABC$ be oriented so that $C$ is at the origin, and $B$ is at the point$(a,0)$. [[Image:Law_of_cosines_proof.jpg]] [[Image:Law_of_cosines_proof.jpg]] - ===Distance Formula=== $distance = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$ $distance = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$ Line 40: Line 48: Since $\cos^{2} C + \sin^{2} C = 1$, Since $\cos^{2} C + \sin^{2} C = 1$, - $c = \sqrt {(a^{2}+b^{2}-2ab \cos C+b^{2}}$ + $c = \sqrt {a^{2}+b^{2}-2ab \cos C+b^{2}}$ Square both sides for Square both sides for - $c^{2} = (a^{2}+b^{2}-2ab \cos C+b^{2}$ + $c^{2} = a^{2}+b^{2}-2ab \cos C+b^{2}$ - ==Example Triangulation== + ==Example Problem== Complete the triangle using the law of cosines. Complete the triangle using the law of cosines.

Revision as of 10:53, 3 June 2011

Law of Cosines

The law of cosines is a trigonometric extension of the Pythagorean Theorem.

Basic Description

The law of cosines is a formula that helps in solving triangles when two or three side lengths of a triangle are known. The formula combines the squares of two side lengths of a triangle and some offset, classified by the cosine of a particular angle, to calculate the square of the third side. For this reason, the law of cosines is often thought of as the generalization of the Pythagorean theorem, which only applies to right triangles. The law of cosines adds an extra term to the Pythagorean theorem so that a third side length of a triangle can be determined when there is no right angle.
$c^{2} = a^{2} + b^{2} - 2ab \cos C$

When to use it: SAS, SSS.

A More Mathematical Explanation

Proof

By the Pythagorean Theorem

An easy way to think of the law of cosines is as an extens [...]

Proof

By the Pythagorean Theorem

An easy way to think of the law of cosines is as an extension of the Pythagorean theorem:

$a^{2} + b^{2} = c^{2}$

Using the Distance Formula

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

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

$c$ 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 Problem

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

$0 = \cos B$

Using inverse trigonometry, we know that

$B = 90^\circ$

And we can find the last angle measure $A$ by subtracting the other two measures from $180^\circ$

$180^\circ - 90^\circ - 45^\circ = 45^\circ$

$A=45^\circ$

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