Question

How to find length of angle bisector in triangle?

Answer 1

Answer:

Since you are given the three sides of the triangle, a, b and c, first you can find the angle 2C which is bisected. You do this using cosine rule -

c2=a2+b2−2abCos(2C)

Cos(2C)=a2+b2−c22bc

Using half angle formula -

Cos(C)=√[1−Cos(2C)2] …… eqn 1

Again using cosine rule, referring to figure we can write -

n2m2=a2+x2−2axCos(C)b2+x2−2bxCos(C)

According to angle bisector rule -

bm=an

Therefore

a2b2=a2+x2−2axCos(C)b2+x2−2bxCos(C)

a2(b2+x2−2bxCos(C))=b2(a2+x2−2axCos(C)

a2b2+a2x2−2a2bxCos(C)=a2b2+b2x2−2ab2xCos(C)

(a2−b2)x2−2ab(a−b)Cos(C)x=0

x[(a2−b2)x−2ab(a−b)Cos(C)]=0

(a2−b2)x−2ab(a−b)Cos(C)=0 … since x cannot be zero

x=2ab(a−b)Cos(C)a2−b2

x=2ab(a−b)Cos(C)(a+b)(a−b)

x=2abCos(C)a+b