Prove that in any triangle ABC,$cos A=\large \frac{b^2+c^2-a^2}{2bc}$,$where \overrightarrow{a},\overrightarrow{b} and \overrightarrow{c}$ are the vectors of sides BC, CA and AB respectively.

Toolbox:
• In a triangle $ABC \;\overrightarrow {AB}+\overrightarrow {BC}+\overrightarrow {CA}=0$
• $\overrightarrow{a}.\overrightarrow{b}=|a| |b| \cos \theta$
To prove $cos A=\large \frac{b^2+c^2-a^2}{2bc}$
We know that in $\Delta$ ABC $\overrightarrow {AB}+\overrightarrow {BC}+\overrightarrow {CA}=0$
Let $\overrightarrow{BC}=\overrightarrow a, \: \overrightarrow{CA}=\overrightarrow b, \: \overrightarrow{AB}=\overrightarrow c$
where $\overrightarrow{BC}=-\bigg[\overrightarrow {AB}+\overrightarrow{CA}\bigg]$
Squaring on both sides we get
$\overrightarrow{BC}. \overrightarrow{BC}=(\overrightarrow {AB}+\overrightarrow{CA}).(\overrightarrow {AB}+\overrightarrow{CA})$
$\overrightarrow{BC}. \overrightarrow{BC}=\overrightarrow {AB}.\overrightarrow{AB}+\overrightarrow {CA}.\overrightarrow{CA}+2 \overrightarrow{AB}. \overrightarrow{CA}$
$a^2=c^2+b^2+2 \overrightarrow c.\overrightarrow b$
$a^2=c^2+b^2+2 |\;c\;|\;|\;b\;| \cos (\pi-A)$
But we know $\cos (\pi-A)=-\cos A$
Therefore $a^2= b^2+c^2-2bc \cos A$
Therefore $\cos A=\large \frac{b^2+c^2-a^2}{2bc}$
Hence proved
answered May 30, 2013 by