Question #0e494

Question

4338 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-23T15:21:26

For $DeltaABC$

we have $A+B+C=pi$

So $cot(A+B)=cot(pi-C)$

$=>(cotAcotB-1)/(cotB+cotA)=-cotC$

$=>cotAcotB+cotBcotC+cotCcotA=1" "color(red)([1])$

Now it is given

$cotA+cotB+cotC=sqrt3$

Squaring both sides we get

$cot^2A+cot^2B+cot^2C+2(cotAcotB+cotBcotC+cotCcotA)=3" "color(red)([2])$

Now subtracting thrice of [1] from [2] we get

$cot^2A+cot^2B+cot^2C-(cotAcotB+cotBcotC+cotCcotA)=0$

$=>2cot^2A+2cot^2B+2cot^2C-2cotAcotB-2cotBcotC-2cotCcotA=0$

$=>(cotA-cotB)^2+(cotB-cotC)^2+(cotC-cotA)^2=0$

Now sum of three squared terms each of which is positive being zero each will be equal to zero.

Hence

$cotA=cotB=cotC$

$=>A=B=C$

This implies that $DeltaABC$ is equilateral.