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Difference between revisions of "Law of Sines"

Revision as of 06:23, 7 October 2007

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, $\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}=2R$ .

Contents

1 Proof
- 1.1 Method 1
- 1.2 Method 2
2 See also

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.

This picture could be replaced by an asymptote drawing. It would be appreciated if you do this.

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: $[ABC] = \frac{1}{2}ab\sin C$

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

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

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

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

See also

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