Question

Guys i need the complete solution of sec-b last sum.......17 pls dont dare spam ,........

Answer 1

Answer:

We have been given the expression

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

Now,

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

=>

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

Which is,

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

So,

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

or,

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

But from Cosine Rule,

LHS is cos C

so,

cos C = 1/2

That means

Angle C = 60°

Since, only cos 60° = 1/2 and C must be within 0 to 180°

Hope this helps you !