Hi can anybody show me how to find x?
-180= -90 - tan^-1 (x/16) - tan^-1 (x/200)
$\displaystyle -180=-90-tan^{-1}(\frac{x}{16})-tan^{-1}(\frac{x}{200})$
Get rid of the negatives by adding the 90 to both sides and then multiplying by -1:
$\displaystyle 90=tan^{-1}(\frac{x}{16})+tan^{-1}(\frac{x}{200})$
Take the derivatives:
$\displaystyle \frac{16}{x^{2}+256}+\frac{200}{x^{2}+40,000}=0$
Now, when we solve this we will get a complex solution, use the real part.
Solving we get $\displaystyle x=40\sqrt{2}i, \;\ x = -40\sqrt{2}i$
The $\displaystyle 40\sqrt{2}$ is a solution.
$\displaystyle -90-tan^{-1}(\frac{40\sqrt{2}}{16})-tan^{-1}(\frac{40\sqrt{2}}{200})=-180$
Hello, al2308!
Solve for $\displaystyle x\!:\;\;-180\;= \;-90 - \tan^{\text{-1}}\!\left(\frac{x}{16}\right) - \tan^{\text{-1}}\!\left(\frac{x}{200}\right)$
We have: . $\displaystyle \tan^{\text{-1}}\!\left(\frac{x}{16}\right) + \tan^{\text{-1}}\!\left(\frac{x}{200}\right) \;=\;90$
Take the tangent of both sides: . $\displaystyle \tan\left[\tan^{\text{-1}}\left(\frac{x}{16}\right) + \tan^{\text{-1}}\left(\frac{x}{200}\right)\right] \;=\;\tan90$
. . $\displaystyle \frac{\tan\left[\tan^{-1}\left(\dfrac{x}{16}\right)\right] + \tan\left[\tan^{-1}\left(\dfrac{x}{200}\right)\right]} {1 - \tan\left[\tan^{-1}\left(\dfrac{x}{16}\right)\right]\left[\tan^{-1}\left(\dfrac{x}{200}\right)\right]}$ .$\displaystyle =\;\tan90 \quad\Rightarrow\quad \frac{\dfrac{x}{16} + \dfrac{x}{200}}{1 - \left(\dfrac{x}{16}\right)\left(\dfrac{x}{200}\rig ht)} \;=\;\tan90$
And we have: . $\displaystyle \frac{\dfrac{x}{16} + \dfrac{x}{200}}{1 - \dfrac{x^2}{3200}} \;=\;\tan90$
Note: the right side is undefined.
Hence, on the left side, the denominator is equal to zero.
. . $\displaystyle 1-\frac{x^2}{3200} \:=\:0\quad\Rightarrow\quad x^2 \:=\:3200 \quad\Rightarrow\quad x \:=\:\pm40\sqrt{2}$
[Check to see if any roots are extraneous.]
Think of it geometrically. If you have a right-angled triangle in which the angles (apart from the right angle ) are α and β, then α + β = 90°. But tan(β) = 1/tan(α) (because if one of them is "opposite over adjacent" then the other one is "adjacent over opposite"). In fact, this formula holds not just for acute angles, but for any values of α and β with α + β = 90°.
So if $\displaystyle \tan^{-1}\Bigl(\frac x{16}\Bigr) + \tan^{-1}\Bigl(\frac x{200}\Bigr) = 90^{\circ}$ then $\displaystyle \frac x{200} = 1\left/\frac x{16}\right. = \frac{16}x$. Thus $\displaystyle x^2 = 16\times200 = 3200$, and $\displaystyle x = \pm40\sqrt2$, as Soroban says.
That's a good question, Moo. I never bothered to look at why it worked. I will have to do that and get back. Okey-doke.
But then again, perhaps it isn't viable. But I know that it happens to work with problems of this type.
I even made up various other problems of this type only using values other than 16 and 200.
If we do the same thing with, say, $\displaystyle tan^{-1}(\frac{x}{22})+tan^{-1}(\frac{x}{168})=90$
It works also. Upon taking the derivatives and doing what we done previously, we get the solutions $\displaystyle \pm{4\sqrt{231}}i$.
The real parts satisfy the equation. $\displaystyle tan^{-1}(\frac{4\sqrt{231}}{22})+tan^{-1}(\frac{4\sqrt{231}}{168})=90$
I tried it with many others and it always seems to pan out. I stumbled upon this method myself, but never bothered trying to figure out why it works. May be something rather obvious. Maybe someone else has an idea. I will look at it over the weekend.
Try another.
$\displaystyle tan^{-1}(x/4)+tan^{-1}(x/5)=90$
Derivative of $\displaystyle tan^{-1}(x/4)=\frac{4}{x^{2}+16}$
Derivative of $\displaystyle tan^{-1}(x/5)=\frac{5}{x^{2}+25}$
$\displaystyle \frac{4}{x^{2}+16}+\frac{5}{x^{2}+25}=0$
$\displaystyle x=\pm{2\sqrt{5}}i$
$\displaystyle tan^{-1}(\frac{2\sqrt{5}}{4})+tan^{-1}(\frac{2\sqrt{5}}{5})=90$
When I tried this I was thinking along the lines of the derivatives being equal at the same point, then this point subbed into the arctans and summing to 90, though I did not try to prove it in a general case.
If we just set the derivatives equal and solved for x:
$\displaystyle \frac{16}{x^{2}+256}=\frac{200}{x^{2}+40,000}$
$\displaystyle x=\pm{40\sqrt{2}}$
The derivatives have equal slope at the points $\displaystyle \pm{40\sqrt{2}}$
You really do not need to mess with the complex.
These values will result in the the two angles summing to 90 degrees.