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 ABC , with corresponding side lengths a,b a, b and c c and corresponding angles α,β \alpha, \beta and γ \gamma, and R R as the radius of the circumcircle of triangle ABC ABC, we have the following:

asinα=bsinβ=csinγ=2R. \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 O be the center of the circumcircle. Let D D be the midpoint of BC BC, then OD OD is perpendicular to BC BC. Note that BOC \angle BOC is equal to 2α 2\alpha or 3602α 360^\circ - 2\alpha, depending on whether O O is in the triangle or not. This gives BOD=α \angle BOD = \alpha or 180α 180^\circ - \alpha, and thus sinBOD=sinα \sin BOD = \sin \alpha. As such, BDOB=sinαasinα=2R. \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 ABC given some of its sides and angles. It is worthwhile to mention the 'ambiguous case': Given BC=a,AB=c,BAC=α BC=a, AB=c, \angle BAC = \alpha such that csinα<a<c 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 ABC is equal to abc4R. \frac {abc} {4R}.

Let D D be the foot of the perpendicular from A A to BC BC. Using BC BC as the base and AD AD as the height, the area of the triangle is 12aAD \frac {1}{2} a \cdot |AD|. From right triangle CAD CAD, sinγ=ADb \sin \gamma = \frac {|AD|} {b}. Thus, the area of the triangle is 12aAD=12absinγ \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, csinγ=2R \frac {c}{\sin \gamma} = 2R, thus the area of the triangle is 12absinγ=12abc2R=abc4R \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 AB and CD CD are drawn, such that they intersect at P P within the circle. What is the value of AP2+BP2+CP2+DP2? AP^2 + BP^2 + CP^2 +DP^2?

Applying the Pythagorean theorem to triangles APC APC and BPD BPD, we get AP2+CP2=AC2 AP^2 + CP^2 = AC^2, BP2+PD2=BD2 BP^2 + PD^2 = BD^2. Applying the Extended Sin Rule to triangles ABC ABC and BCD BCD, we get AC=2RsinABC AC = 2R \sin \angle ABC and BD=2RsinDCB BD = 2R \sin \angle DCB, where R=5 R=5 is the radius of the circle. Applying the trigonometic form of the Pythagorean theorem to triangle PBC PBC, we get sin2PCB+sin2PBC=1 \sin^2 \angle PCB + \sin ^2 \angle PBC = 1. Putting this all together, we have

AP2+BP2+CP2+DP2=AC2+BD2=100sin2ABC+100sin2BCD=100(sin2PBC+sin2PCB)=100. \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 P.


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

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

ABsinADB=BDsinBADABBD=sinADBsinBAD. \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 ACD, we get:

ACsinADC=CDsinDACACCD=sinADCsinDAC. \frac {AC}{\sin \angle ADC} = \frac {CD}{\sin\angle DAC} \Rightarrow \frac {AC}{CD} = \frac {\sin \angle ADC} {\sin \angle DAC}.

Since AD AD is the angle bisector of BAC \angle BAC, BAD=DAC \angle BAD = \angle DAC implies that sinBAD=sinDAC \sin \angle BAD = \sin \angle DAC. Since BDC BDC is a straight line, ADB=180ADCsinADB=sinADC \angle ADB = 180^\circ - \angle ADC \Rightarrow \sin \angle ADB = \sin \angle ADC. Thus, ABBD=sinADBsinBAD=sinADCsinDAC=ACCD. \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
5 years, 3 months ago

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

Tarun Kumar - 4 years, 11 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 - 4 years, 4 months ago

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