# Find the cosine of angle $A$ if the sides of triangle $ABC$ are related as $3:2:2$.

So it has been a while since I last studied Trigonometry, so I thought I should revise it and solve on it. So I was solving on Cosine/Sine laws and I came across this question that I really couldn't figure out. Help would be greatly appreciated.

In triangle ABC, if a:b:c = 3:2:2, then cos A equals ...

Informative explanations for how to solve this kind of problems when I run into them again is what I am really looking for, thanks in advance.

• As an aside, the triangle whose sides form the ratio $(1:2:2)$ has the property that all four triangle centers are not only co-linear, but equally distanced. – Lucian Oct 23 '15 at 4:07

Since $a:b:c=3:2:2$ we have $\frac{a}{b}=\frac{3}{2}$ and $\frac{b}{c}=\frac{2}{2}=1,\quad$ then $\quad b=c=\frac{2a}{3}$.

Now, from the Cosine Law we get $$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{\left(\frac{2a}{3}\right)^2+\left(\frac{2a}{3}\right)^2-a^2}{2\left(\frac{2a}{3}\right)^2}=\frac{\frac{8}{9}-1}{\frac{8}{9}}=-\frac{1}{8}$$

• Can you explain further? I mean, I know the Cosine Law, but I can't understand how we got that fraction (2a/3). Thank you! – Eyad H. Oct 23 '15 at 1:18
• Were you notified with my comment? I hope you were. ;-; – Eyad H. Oct 23 '15 at 1:54