Solving an equation with arctan and arcsin I am trying to do what I think a problem with a simple answer. Here are the two equations I have resolved the problem down to: 
$$\angle A = \arctan \frac{28}{x}$$ 
and $$\angle A = \arcsin \frac{1}{12-x}$$  Is it not right then that I can make these equal to each other and I have only one variable. But I do not have the math skills to solve it....help please.
 A: Hint.  You have
$$\tan A=\frac{28}{x}\ ,\quad \sin A=\frac{1}{12-x}\ .$$
From $\tan A$ and $\sin A$ you can find $\cos A$, and from $\cos^2A+\sin^2A=1$ you can get a quadratic equation for $x$.
Good luck!
A: If $$\tan A = \frac{28}{x},\qquad \sin A = \frac{1}{12-x} = \frac{28}{28(12-x)}$$
then $A$ equals $\widehat{A}$ in the the triangle $ABC$, having $AB\perp BC$, $AB=x$, $BC=28$, $AC=28(12-x)$. Now the Pythagorean theorem gives:
$$\left(28(12-x)\right)^2 = 28^2 + x^2, $$
or:
$$ 783 x^2 - 18816 x + 112112 = 0,$$
hence:

$$ x=\frac{28}{261}\left(112\pm\sqrt{103}\right).$$

A: The picture below illustrates your situation, I think. The grey thing with the black border is your piece of tubing. The picture is not to scale.

The two pink triangles are similar, so 
$$
\frac{h}{28} = \frac{12-x}{1}
$$
But, by Pythagoras, $h = \sqrt{28^2 + x^2}$, so we get
$$
28(12-x) = \sqrt{28^2 + x^2}
$$
Squaring both sides and rearranging gives
$$
783 x^2 - 18816 x + 112112 = 0
$$
You can solve this equation with the quadratic formula. You get
$$
x = 10.926555694166351 \quad \text{or} \quad x = 13.104095646829817 
$$
The first solution $(x=10.926555...)$ is the one we want. Then $\sin A = 1/(12-x) = 1/1.0734443 = 0.93158$, and so $A = \sin^{-1}(0.93158) = 68.68^\circ$.
A: Note 
$$1+\cot^2A=\csc^2A, \cot A=\frac{1}{\tan A}, \csc A=\frac{1}{\sin A} $$
and now I think you can get an equation to solve $x$.
A: $\displaystyle\angle A = \arctan\frac{28}x  = \arcsin\frac1{12-x}$
Using the definition of Principal values, $\displaystyle-\frac\pi2\le\angle A\le\frac\pi2$
Case $\#1:$ If $\displaystyle0\le\angle A\le\frac\pi2, \frac{28}x\ge0\iff x>0$ and $\displaystyle\frac1{12-x}\ge0\iff 12-x>0\iff x<12$  
$\displaystyle\implies 0<x<12$
Case $\#2:$ If $\displaystyle-\frac\pi2\le\angle A<0, \frac{28}x<0\iff x<0$ and $\displaystyle\frac1{12-x}<0\iff 12-x<0\iff x>12$  
$\displaystyle\implies 0>x>12$ which is impossible
Now that $\displaystyle\tan A=\frac{28}x,\sin A=\frac1{12-x}$
and $\displaystyle\csc^2A-\cot^2A=1\iff\frac1{\sin^2A}-\frac1{\tan^2A}=1$
Hope you can take it home from here
