# Trigonometric Trianle

• November 28th 2006, 06:56 AM
camherokid
Trigonometric Trianle
what is the value of X when Q is the biggest?
• November 28th 2006, 07:02 AM
ThePerfectHacker
What is Q?
(You cetainly do not mean the Q continuum).
• November 28th 2006, 10:40 PM
earboth
Quote:

Originally Posted by camherokid
what is the value of X when Q is the biggest?

Hello,

I assume that you want to know for which value of x has the angle $\theta$ a maximum.

$\theta(x)=\arctan\left(\frac{420}{x}\right)-\arctan\left(\frac{380}{x}\right)$

You'll get an extremum if the first derivative of this function equals zero:

$\theta'(x)=\frac{1}{1+\left(\frac{420}{x} \right)^2}\cdot \left(\frac{-420}{x}\right)-\frac{1}{1+\left(\frac{380}{x} \right)^2}\cdot \left(\frac{-380}{x}\right)$

After a few and simple :D transformations you'll get:

$\theta'(x)=\frac{40·(159600 - x^2)}{(x^2 + 144400)·(x^2 + 176400)}$

That means: $\theta'(x)=0\text{ if }159600 - x^2=0$. Solve for x. I've got:
$x= 399.4996... \approx399.5$

There is of course a negative solution of x but thisvalue isn't very realistic with your problem.

EB
• November 29th 2006, 05:14 AM
camherokid
Thanks you Earboth...for the solution.
I knew how to get to the Tan formula...
but I just did not what to do with it.
$\theta$ is the angle...but I wrote Q. I don't know
where I can find that sign.

Great Job...!
• November 29th 2006, 06:02 AM
earboth
Quote:

Originally Posted by camherokid
...
$\theta$ is the angle...but I wrote Q. I don't know
where I can find that sign.

Hello,

maybe this will help a little bit further: In this forum you can find a link to the LaTeX-tutorial.