Question

if in ∆ABC 1/a+c + 1/b+c =3/abc then prooved thar C =60°

Answer 1

hey !!!

here is ur solution :-

from question .

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

•°•b+c+ a+c/(a+c)(b+c) = 3/a+b+c

or, a+b+2c /(a+c)(b+c) =3/a+b+c

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

or, a² +ab +ac +ba +b² +bc +2ca +2cb +2c² =3(ab+ ac+bc +c² )

or , a² +b² +2c² +2ab +3bc +3ca =3ab +3ac +3c²

or , a² + b² -c² = ab

or , a² +b² -c² /2ab = ab/2ab

or, cosC = 1/2

cosC = cos60°

C =60° prooved...

hope it helps !!

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