Difference between revisions of "Law of Sines"

Revision as of 11:52, 2 February 2008

The Law of Sines is a useful identity in a triangle, which, along with the law of cosines and the law of tangents can be used to determine sides and angles. The law of sines can also be used to determine the circumradius, another useful function.

Theorem

In triangle $\triangle ABC$ , where $a$ is the side opposite $A$ , $b$ opposite $B$ , $c$ opposite $C$ , and where $R$ is the circumradius:

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

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.

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 $\frac{2}{abc}$ yields:

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

Problems

Introductory

If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?

$\mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 8/\sqrt{15} \qquad \mathrm{(C) \ } 5/2 \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2$

(Source)

Intermediate

Triangle $ABC$ has sides $\overline{AB}$ , $\overline{BC}$ , and $\overline{CA}$ of length 43, 13, and 48, respectively. Let $\omega$ be the circle circumscribed around $\triangle ABC$ and let $D$ be the intersection of $\omega$ and the perpendicular bisector of $\overline{AC}$ that is not on the same side of $\overline{AC}$ as $B$ . The length of $\overline{AD}$ can be expressed as $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find the greatest integer less than or equal to $m + \sqrt{n}$ .

(Source)

Advanced

Let $ABCD$ be a convex quadrilateral with $AB=BC=CD$ , $AC \neqBD$ (Error compiling LaTeX. Unknown error_msg), and let $E$ be the intersection point of its diagonals. Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120^{\circ}$ .

(Source)

@@ Line 2: / Line 2: @@
 ==Theorem==
-In triangle <math>\triangle ABC</math>, where <math>a</math> is the side opposite <math>A</math>, <math>b</math> opposite <math>B</math>, and <math>c</math> opposite <math>C</math>:
+In triangle <math>\triangle ABC</math>, where <math>a</math> is the side opposite <math>A</math>, <math>b</math> opposite <math>B</math>, <math>c</math> opposite <math>C</math>, and where <math>R</math> is the circumradius:
 <math>\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R</math>

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Difference between revisions of "Law of Sines"

Revision as of 11:52, 2 February 2008

Contents

Theorem

Proof

Method 1

Method 2

Problems

Introductory

Intermediate

Advanced

See Also