Question

The vertices of a triangle ABC are A (0, 0), B (2, −1) and C (9, 2). Find cos B.

Answer 1

We know that $\cos B=\frac{a^2+c^2-b^2}{2ac}$, where a = BC, b = CA and c = AB are the lengths of the sides of $∆$ABC.
Thus,
$a=BC=\sqrt{{\left(2-9\right)}^2+{\left(-1-2\right)}^2}=\sqrt{49+9}=\sqrt{58}$
$b=AC=\sqrt{{\left(0-9\right)}^2+{\left(0-2\right)}^2}=\sqrt{81+4}=\sqrt{85}$
$c=AB=\sqrt{{\left(2-0\right)}^2+{\left(-1-0\right)}^2}=\sqrt{4+1}=\sqrt{5}$

Using cosine formula in $∆ABC$, we get:
$\cos B=\frac{a^2+c^2-b^2}{2ac}$

$\Rightarrow \cos B=\frac{58+5-85}{2\sqrt{58}\times \sqrt{5}}=-\frac{11}{\sqrt{290}}$