#sinerule

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Proof of Sine Rule and Cosine Rule of the triangle

Let, In a △ABC, the length of sides: BC=a, CA=b, and AB=c then according to the Sine Formula

a/Sin A =b/Sin B =c/Sin C =2R

Here R is the Circumradius of the △ABC.

Here is the proof of this formula.

Draw AD⟂BC, the height AD=h

In △BDA, Sin(B)=h/c

h=c.Sin(B)

Similarly, In △CDA

Sin(C)=h/b

h=b.Sin(C)

So, c.Sin(B)=b.Sin(C)

b/Sin(B)=c/Sin(C) ……………(i)

Now, Draw BE⟂AC, the height BE=k

a/Sin(A)=c/Sin(C) ……………(ii)

From the equation (i) and (ii):

a/Sin(A) =b/Sin(B) =c/Sin(C) ……………(iii)

— — — — — — — —

Draw a Circumcircle of △ABC. Here O is the Circumcenter of the triangle.

The length of Circumradius=R

So, the length of the diameter, BD=2R

∠BAC=∠BDC=∠A (Angles, subtended by the same arc at the circumference in the same segment)

∠BCD=90° (Angle made by the diameter at the Circumference)

Now in △BCD, according to Sine Rule:

BC/Sin(BDC)}= BD/Sin(90°)

a/Sin(A) =2R/1

a/Sin(A) =2R …………(iv)

From equation (iii) and (iv):

a/Sin(A) =b/Sin(B) =c/Sin(C)=2R

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