# Sine Law

Sine Law:

Sine law states that in any triangle ABC:

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

And also with some mathematics we can also prove the following :

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

Which is also closely related to the sine law.

Where , a , b and c are sides of a triangle , A , B and C are angles opposite to sides a , b, and c correspondingly and  R is the circum-radius of the triangle as shown in following figure:

sine law

Proof of Sine Law:

Let us consider a Triangle ABC  , placed in the standard position with the vertex A at the origin and side AB along the positive x-axis as in the figure below:

Now,

* The co-ordinates of A are ( 0 , 0 )

* The co-ordinates of B are ( c , 0 )

* The co-ordinates of C are ( b cosA , b sinA )

Now ,

Area of the triangle ABC is:

$\dfrac{1}{2} \times base \times altitude \\ \\ = \dfrac{1}{2} \times AB \times ordinate \, of \, vertex C \\ \\ = \dfrac{1}{2} \times b \times c \times \sin A$

Similarly if we place the other angles B and C in the standard position or origin the we also get the following by similar method:

Area of triangle =

$\dfrac{1}{2} \times a \times c \times \sin B \\ \\ \dfrac{1}{2} \times a \times b \times \sin C$

By combining these three formulas we get:

$\Delta = \dfrac{1}{2} b c \sin A = \dfrac{1}{2} a c \sin B = \dfrac{1}{2} a b \sin C$

Hence:

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

Now , we shall prove the sine law with another approach to find the sine relation or connection between the radius of circum-circle with the sine of angles and their opposite sides.

Or now we shall prove:

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

To prove this we denote the circum-centre of the triangle ABC by “O”.

There might be three cases in which angle “A” is either acute (fig. a) , obtuse (fig. b) or right angled (fig. c).

In any case let us join the circum-centre of the triangle “O” to B and produce it in another side up to “D” as shown in figures blow:

sine law

In first two cases above:

angle BDC = 90 degrees ( Because angle made on circumference from diameter is always 90 degrees)

Thus:

$BC = BD \times \sin BDC$

Or , $a = 2R \times \sin A$

And in the third case it is obvious that : $a = 2R \times \sin A$ because in this case angle A is 90 degrees and sine 90 = 1.

Thus we can now conclude:

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

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