Sine law states that in any triangle ABC:
And also with some mathematics we can also prove the following :
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:
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:
* 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 )
Area of the triangle ABC is:
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 =
By combining these three formulas we get:
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:
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:
In first two cases above:
angle BDC = 90 degrees ( Because angle made on circumference from diameter is always 90 degrees)
And in the third case it is obvious that : because in this case angle A is 90 degrees and sine 90 = 1.
Thus we can now conclude:
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