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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.

jeff
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    Can you show the original problem text please? – Cookie Aug 11 '14 at 00:21
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    No problem text. Real world problem. Trying to determine the correct angle to cut 1 inch square tubing ends to angle that results in the tubing 12 inches wide from side to side (tip to tip) and 28 inches tall. If the tubing was infinitely small this would be easy...:-) – jeff Aug 11 '14 at 00:47

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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!

David
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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:

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

Jack D'Aurizio
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  • thx. I realized I was over thinking a bit. atan and asin were superfluous..... – jeff Aug 11 '14 at 02:49
  • You're welcome. But please upvote or accept my answer as soon as your reputation allows you to, if you found it useful. I do not like to solve problems for nothing. – Jack D'Aurizio Aug 11 '14 at 02:58
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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.

enter image description here

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 $A = \sin^{-1}(0.93158) = 68.68^\circ$.

bubba
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  • New to this site and quite amazed. This is exactly the problem. While I could have just carefully measured this, I wanted to determine the angle precisely using math. Thanks. – jeff Aug 11 '14 at 14:44
  • If/when you cut the tubing, I'd be interested to know if my answer is correct. I think it is, but it's different from the one given by Jack A'Aurizio. – bubba Aug 11 '14 at 22:32
  • I checked my answer in a CAD system, and I'm now pretty sure that it's correct. I haven't bothered to find the flaw in the other answer, but I think there must be one. – bubba Aug 12 '14 at 00:03
  • I found out why my quadratic is different from Jack's. It's because his $x$ and my $x$ represent different things. I changed the definition of my $x$, and I now get the same quadratic that he got. So, no inconsistency. The final answer for the angle $A$ remains the same. Apologies to Jack for suggesting that his solution is erroneous. It's not. – bubba Aug 12 '14 at 00:28
  • I had checked both and I found the roots the same. I did cut the tubing and yes, it was the correct angle. – jeff Aug 12 '14 at 22:40
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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 $x$.

xpaul
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$\displaystyle\angle A = \arctan\frac{28}x = \arcsin\frac1{12-x}$

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

lab bhattacharjee
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