Question

(topic: sol. of triangle)
In a triangle ABC if
a^2+b^2+c^2=ac+ab sqrt(3)
then prove it is right angled. (a,b,c are sides opp. A,B,C)

Answer 1

a^2+b^2+c^2=ac+2ab sqrt(3) /2
a^2+b^2+c^2=ac+2ab cos 30
so we can assume angle C = 30 , then by cos law
c^2=a^2+b^2-2ab cos 30 ,
so , 2ab cos 30 = a^2+b^2 -c^2
substituting this in a^2+b^2+c^2=ac+2ab cos 30
we get,
2c^2 = ac
----> c/a = 1/2
Then from sine Law,
sin A/ sin C = c/a gives us A =90.

Answer 2

neither do i know whether its correct :P
just gave it a try..... if i had known for sure, i would make him reach the answer...

Answer 3

It's not hard to show that if you have a 30-60-90 triangle, then a^2+b^2+c^2 = ac+bc sqrt(3) where c is the hypotenuse.

But going the other way is more difficult.

I can massage things to
\[ a \cos B + b \cos A -\frac{ab}{c} \cos(A+B)= a\cdot \frac{\sqrt{3}}{2} + b \cdot \frac{1}{2} \]
then argue that the coefficients of a and b must match up. Is that a proof?