Inspired by Rakshit Joshi

Geometry Level pending

Consider an arbitrary triangle A B C ABC with ∠ C = 6 0 ∘ \angle C= 60^\circ . Let a , b , c a,b,c denote the sides opposite to the vertices A , B , C A,B,C , respectively.

Find the value of 45 ( 1 a + c + 1 b + c − 3 a + b + c ) 45 \left( \dfrac1{a+c} + \dfrac1{b+c} - \dfrac3{a+b+c} \right) .


The answer is 0.

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A Former Brilliant Member by A Former Brilliant Member

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2 solutions

s= semiperimeter

6 Helpful 0 Interesting 0 Brilliant 0 Confused

I expressed a,b & c as 2RsinA,2RsinB and 2RsinC and arrived at the given ans

0 Helpful 0 Interesting 0 Brilliant 0 Confused

well done chirag that is a good solution

A Former Brilliant Member - 4 years, 7 months ago

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