# Finding tangent for angle of right triangle

Given : Right $\triangle ABC$ $CH$ is altitude and $AH =2$, $BH=8$. Here is drawing

Find the $tan \angle CBA$

Here is how I solved the exercise: From that fact that $ABC$ is right triangle $=>$ $AC^2=AH*AB$ (Altitude on hypotenuse theorem) $=> AC = \sqrt{2*10} = \sqrt{20}$ and $tan \angle CBA = \frac{AC}{AB}=\frac{2*\sqrt{5}}{10}=\frac{\sqrt{5}}{5}$

However the answer in test is $\frac{1}{2}$. I can't find my mistake.

• CH =ah*ab. Not ac – fleablood May 11 '16 at 18:01
• @fleablood no, that's right what he wrote. But he found sin, not tan – openspace May 11 '16 at 18:04
• Oops, I misread ab as hb, ie ch^2 = ah*hb. Which wasn't the intent. I should learn not to post comments in rushed minutes between doctors appointments. – fleablood May 11 '16 at 19:00
• Right. Angle cba = cbh so tan= opp/adj = ca/cb (not ab) or ch/hb. (For some reason, I'm finding triangle chb easier to work with than abc which is we I made the mistaken first post.) – fleablood May 11 '16 at 19:04

As we know : $$CH = \sqrt{AH \cdot BH}$$, Now $$tan(CBA) = \frac{4}{8} = \frac{1}{2}$$
• My I ask why did you prefer to use $CH$ instant of $AC$? – Planet_Earth May 11 '16 at 18:01
• @Planet_Earth it's easier , because of using your values you could evaluate $CH$. Then if you consider CHB triangle you may find your tangent – openspace May 11 '16 at 18:02
• Yeah, it's clear for me how you did it. However both of the answers are provided(together with $\frac{2*\sqrt{5}}{5}$ and 2), I thought that there is some logic behind this(using $CH$ instant of $AC$) – Planet_Earth May 11 '16 at 18:05
• If you want to use AC you can. But tan = CA/CB instead of the CA/AB you wrote. (I believe the triangle being upside down temporarily threw you off; It threw me off briefly.) AC = $\sqrt 20$ so CH = 4 (20- 4 = 16 = $CH^2$) so CB =$\sqrt 80$ ($8^2 + 4^2 = 80$) and AC/CB = 1/2. – fleablood May 11 '16 at 20:18