Question

In a triangle ABC if 1 by a + C + 1 by b + c is equals to 3 by a + b + c then prove that angle C is equals to 60 degrees ​

Answer 1

Step-by-step explanation:

1/(a+c) + 1/(b+c) = 3 / ( a+b+c )

=> (a+b+2c) / (a+c) (b+c) = 3 / (a+b+c)

=> (a+b+2c) (a+b+c) = 3(a+c) (b+c)

=> a^2 + b^2 + 2ab + 2c^2 + 3ac + 3bc = 3ab + 3ac + 3bc + 3c^2

=> a^2 + b^2 + 2ab - 3ab = 3 c^2 - 2 c^2 (cancel 3ac & 3bc)

Now a^2+b^2-ab-c^2=0

a^2+b^2-c^2=ab

a^2+b^2-c^2/ab=0

Divided both side by ½

a^2+b^2-c^2/2ab=1/2

CosC=1/2

Cos C=cos 60

C=60 H. P