Question

In triangle ABC, if (1/b+c) + (1/a+c) = 3/a+b+c then prove that angle C = 60 degree​

Answer 1

Answer:

we know that

a^2 = b^2 + c^2 − 2bc Cos A

b^2 = c^2 + a^2 − 2ca Cos B

c^2 = a^2 + b^2 − 2ab Cos C

angle C given = 60

So,

c^2 = a^2 + b^2 − 2ab cosC

c^2 = a^2 + b^2 − 2ab cos60

c^2 = a^2 + b^2 − ab

Simplify :-

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

= a^2 + b^2 – ab = c^2

Hence Proved ..

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