Question

I have a question again

Answer 1

In Triangle ABC, given 1/(a+c) +1/(b+c) = 3/(a+b+c)
what is the value of angle C

Answer 2

are a,b and c the angles we need to find?

Answer 3

from the pb yes...

Answer 4

\[\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3\]
\[\frac{ab+b^2+bc+ac+bc+c^2}{(a+c)(b+c)}+\frac{a^2+ab+ac+ac+bc+c^2}{(a+c)(b+c)}=3\]
\[2ab+b^2+3bc+3ac+2c^2=3(ab+ac+bc+c^2)\]\[2ab+b^2+3bc+3ac+2c^2=3ab+3ac+3bc+3c^2)\]\[-ab+b^2+c^2=0\]
\[c^2=ab-b^2\]\[c=\pm \sqrt{ab-b^2}\]

Answer 5

that all I could get, hope it helps

Answer 6

thx I'll figure it out. i think the key is wrong

Answer 7

Zed, that is an awesomely formatted explanation. Exquisite!