0
$\begingroup$

Given : Right $\triangle ABC$ $CH$ is altitude and $AH =2$, $BH=8$. Here is drawingenter image description here

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.

$\endgroup$
4
  • $\begingroup$ CH =ah*ab. Not ac $\endgroup$ Commented May 11, 2016 at 18:01
  • 1
    $\begingroup$ @fleablood no, that's right what he wrote. But he found sin, not tan $\endgroup$ Commented May 11, 2016 at 18:04
  • $\begingroup$ 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. $\endgroup$ Commented May 11, 2016 at 19:00
  • $\begingroup$ 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.) $\endgroup$ Commented May 11, 2016 at 19:04

1 Answer 1

Reset to default
0
$\begingroup$

As we know : $$CH = \sqrt{AH \cdot BH}$$, Now $$tan(CBA) = \frac{4}{8} = \frac{1}{2}$$

$\endgroup$
6
  • $\begingroup$ My I ask why did you prefer to use $CH$ instant of $AC$? $\endgroup$ Commented May 11, 2016 at 18:01
  • $\begingroup$ @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 $\endgroup$ Commented May 11, 2016 at 18:02
  • $\begingroup$ 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 $AC$) $\endgroup$ Commented May 11, 2016 at 18:05
  • $\begingroup$ @Planet_Earth actually I thought that it's easier $\endgroup$ Commented May 11, 2016 at 18:07
  • $\begingroup$ 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. $\endgroup$ Commented May 11, 2016 at 20:18

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.