The Law of Sines is a relationship linking the sides of a triangle with the sine of their corresponding angles. The statement is as follows: Given triangle $ABC$, with corresponding side lengths $a, b$ and $c$ and corresponding angles $\alpha, \beta$ and $\gamma$, and $R$ as the radius of the circumcircle of triangle $ABC$, we have the following:

$\frac { a}{ \sin \alpha} = \frac {b}{\sin \beta} = \frac {c} {\sin \gamma} = 2R.$

**Note:** The statement without the third equality is often referred to as the Sine Rule. The third equality is often referred to as the Extended Sine Rule.

## Proof of the Extended Sine Rule

Let $O$ be the center of the circumcircle. Let $D$ be the midpoint of $BC$, then $OD$ is perpendicular to $BC$. Note that $\angle BOC$ is equal to $2\alpha$ or $360^\circ - 2\alpha$, depending on whether $O$ is in the triangle or not. This gives $\angle BOD = \alpha$ or $180^\circ - \alpha$, and thus $\sin BOD = \sin \alpha$. As such, $\frac {BD}{OB} = \sin \alpha \Rightarrow \frac {a} {\sin \alpha } = 2R.\, _\square$

One real-life application of the sine rule is the Sine Bar, which is used to measure the angle of a tilt in engineering. Other common examples include measurement of distances in navigation, and the measurement of distance between two stars in astronomy.

A common textbook application of the sine rule, is to determine the triangle $ABC$ given some of its sides and angles. It is worthwhile to mention the 'ambiguous case': Given $BC=a, AB=c, \angle BAC = \alpha$ such that $c \sin \alpha < a < c$, then there are 2 distinct triangles that satisfy such a configuration.

## Show that the area of triangle $ABC$ is equal to $\frac {abc} {4R}.$

Let $D$ be the foot of the perpendicular from $A$ to $BC$. Using $BC$ as the base and $AD$ as the height, the area of the triangle is $\frac {1}{2} a \cdot |AD|$. From right triangle $CAD$, $\sin \gamma = \frac {|AD|} {b}$. Thus, the area of the triangle is $\frac {1}{2} a \cdot |AD| = \frac {1}{2} a b \sin \gamma$, which is often quoted. Now, using the Extended sine rule, we have, $\frac {c}{\sin \gamma} = 2R$, thus the area of the triangle is $\frac {1}{2} a b \sin \gamma = \frac {1}{2} a b \frac {c}{2R} = \frac {abc} {4R}$. $_\square$

## In a circle of radius 5, two perpendicular chords $AB$ and $CD$ are drawn, such that they intersect at $P$ within the circle. What is the value of $AP^2 + BP^2 + CP^2 +DP^2?$

Applying the Pythagorean theorem to triangles $APC$ and $BPD$, we get $AP^2 + CP^2 = AC^2$, $BP^2 + PD^2 = BD^2$. Applying the Extended Sin Rule to triangles $ABC$ and $BCD$, we get $AC = 2R \sin \angle ABC$ and $BD = 2R \sin \angle DCB$, where $R=5$ is the radius of the circle. Applying the trigonometic form of the Pythagorean theorem to triangle $PBC$, we get $\sin^2 \angle PCB + \sin ^2 \angle PBC = 1$. Putting this all together, we have

$\begin{aligned} AP^2 + BP^2 + CP^2 + DP^2 &= AC^2 + BD^2 \\ &= 100 \sin ^2 \angle ABC + 100 \sin^2 \angle BCD \\ &= 100 (\sin^2 \angle PBC + \sin^2 \angle PCB) \\ &=100. _\square \end{aligned}$

Note that this sum is independent of the point $P$.

## [Angle Bisector Theorem] In triangle $ABC$, let $D$ be a point on $BC$ such that $AD$ is the angle bisector of $\angle BAC$. Show that $\frac {AB}{BD} = \frac {AC}{CD}$.

Applying the sine rule to triangle $ABD$, we get:

$\frac {AB}{\sin \angle ADB} = \frac {BD} {\sin \angle BAD} \Rightarrow \frac {AB}{BD} = \frac {\sin \angle ADB}{\sin \angle BAD}.$

Applying the sine rule to triangle $ACD$, we get:

$\frac {AC}{\sin \angle ADC} = \frac {CD}{\sin\angle DAC} \Rightarrow \frac {AC}{CD} = \frac {\sin \angle ADC} {\sin \angle DAC}.$

Since $AD$ is the angle bisector of $\angle BAC$, $\angle BAD = \angle DAC$ implies that $\sin \angle BAD = \sin \angle DAC$. Since $BDC$ is a straight line, $\angle ADB = 180^\circ - \angle ADC \Rightarrow \sin \angle ADB = \sin \angle ADC$. Thus, $\frac {AB}{BD} =\frac {\sin \angle ADB}{\sin \angle BAD} = \frac {\sin \angle ADC} {\sin \angle DAC} = \frac {AC}{CD}. _\square$

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TopNewestwe are presently studying this topic in the school...I proved this while doing vectors

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A regular pentagon is inscribed in a circle whose radius is 5 cm. Find the one of the sides of the pentagon by using law of sines.

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