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Law of Sines


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.

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
2 years, 7 months ago

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we are presently studying this topic in the school...I proved this while doing vectors Tarun Kumar · 2 years, 3 months ago

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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. Rondarski Roe · 1 year, 8 months ago

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