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# Law of Sines

## Definition

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, $\frac {BD}{OB} = \sin \alpha \Rightarrow \frac {a} {\sin \alpha } = 2R.\, _\square$

## Technique

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.

## Application and Extensions

### 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{align} 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{align}

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, $\frac {AB}{BD} =\frac {\sin \angle ADB}{\sin \angle BAD} = \frac {\sin \angle ADC} {\sin \angle DAC} = \frac {AC}{CD}. _\square$

Note by Arron Kau
3 years ago

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we are presently studying this topic in the school...I proved this while doing vectors · 2 years, 8 months ago