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