Question

The vertices of a triangle are $\text{A(1,1),B(4,5)}$ and $\text{C(6,13)}$. Find $\text{cosA}$.

Answer 1

Step 1: Use distance formula and find the length of each sides of the triangle

First we find the value of $\text{a,b,c}$.

In $△\text{ABC}$ we know that $\text{a = BC, b = CA, c = AB}$

We find $\text{a,b,c}$ by using distance formula

Distance formula : We find the distance of two points ${\text{A(x}}_1{\text{,y}}_1\text{)}$ and point ${\text{B(x}}_2{\text{,y}}_2\text{)}$ like this

$\text{AB=}\sqrt{{{({\text{x}}_2{\text{-x}}_1\text{)}}^2{\text{+(y}}_2{\text{-y}}_1\text{)}}^2}$

$\text{a=BC=}\sqrt{{(6-4)}^2+{(13-5)}^2}\text{=}\sqrt{2^2+8^2}\text{=}\sqrt{4+64}\text{=}\sqrt{68}$

$\text{b=CA=}\sqrt{{(6-1)}^2+{(13-1)}^2}\text{=}\sqrt{5^2+{12}^2}\text{=}\sqrt{25+144}\text{=}\sqrt{169}\text{=13}$

$\text{c=AB=}\sqrt{{(4-1)}^2+{(5-1)}^2}\text{=}\sqrt{3^2+4^2}\text{=}\sqrt{9+16}\text{=}\sqrt{25}\text{=5}$

Step 2: Use cosine rule

We have a formula to find $\text{cosA=}\frac{{\text{b}}^2{\text{+c}}^2{\text{-a}}^2}{2\text{bc}}$

Now we put the values of $a,b$ and $c$ in the above formula

$\begin{array}{rcl}\cos A& =& \frac{{13}^2+5^2-{\left(\sqrt{68}\right)}^2}{2\times 13\times 5}\\ & =& \frac{169+25-68}{130}\\ & =& \frac{126}{130}\end{array}$

$=\frac{63}{65}$

Hence, $\cos A=\frac{63}{65}$