Difference between revisions of "Law of Sines"

Revision as of 11:55, 10 July 2006

Given a triangle with sides of length a, b and c, opposite angles of measure A, B and C, respectively, and a circumcircle with radius R, $\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}=2R$ .

Proof

Method 1

In the diagram below, circle $O$ circumscribes triangle $ABC$ . $OD$ is perpendicular to $BC$ . Since $\triangle ODB \cong \triangle ODC$ , $BD = CD = \frac a2$ and $\angle BOD = \angle COD$ . But $\angle BAC = 2\angle BOC$ making $\angle BOD = \angle COD = \theta$ . Therefore, we can use simple trig in right triangle $BOD$ to find that

$\sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R.$

The same holds for b and c thus establishing the identity.

Method 2

This method only works to prove the regular (and not extended) Law of Sines.

The formula for the area of a triangle is: $\displaystyle [ABC] = \frac{1}{2}ab\sin C$

Since it doesn't matter which sides are chosen as $a$ , $b$ , and $c$ , the following equality holds:

$\displaystyle \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C$

Multiplying the equation by $\displaystyle \frac{2}{abc}$ yeilds:

$\displaystyle \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

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Difference between revisions of "Law of Sines"

Revision as of 11:55, 10 July 2006

Contents

Proof

Method 1

Method 2

See also

Revision as of 21:11, 29 June 2006 (view source) Joml88 (talk \| contribs) ← Older edit	Revision as of 11:55, 10 July 2006 (view source) JBL (talk \| contribs) m (Law of sines moved to Law of Sines: consistency (I hope)) Newer edit →
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