Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
0 votes

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.

Can you answer this question?

1 Answer

0 votes
  • 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 meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App