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
-----------------
Read this article for complete details
Some very useful Trigonometry formulas and Identities with proof are given here. Here we would also discuss some of the very important pract
Join us on Telegram at ‘Logicxonomy’ for regular updates









