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:

\[\cos B = \frac{a^2 + c^2 - b^2}{2ac}\]

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