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I was never good in trigonometry. I have a rectangle with dimensions $L_1$ and $W_1$. I want to rotate it so that it fits inside another rectangle with dimensions $L_2$ and $W_2$. I need to find the angle.

I have worked out the formula that I need is: $$L_1 \cos\theta + W_1 \sin \theta = L_2$$

I don't know how to solve for $\theta$.

I understand that there won't be a solution for $$L_2 > \sqrt{L_1^2 + W_1^2}$$

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MJD
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LeppyR64
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  • Use $a\cos \phi +b\sin \phi =\sqrt{a^2+b^2}\sin \left ( \phi +\psi \right )$ where $\psi =\arctan \frac{b}{a}$ – EricAm Aug 26 '14 at 13:35
  • Any questions, Jason, about the answers that have been provided? – Gerry Myerson Aug 27 '14 at 23:39
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    Both answers make sense in my head. Just haven't got back to my paper yet to make sure I grasp them 100%. Thanks for providing the answers and following up! I plan to close them off when I return home tomorrow. – LeppyR64 Aug 28 '14 at 00:26

2 Answers2

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Define $Q=\sqrt{L_1^2+W_1^2}$. Divide your equation by $Q$; $${L_1\over Q}\cos\theta+{W_1\over Q}\sin\theta={L_2\over Q}$$ Define $\phi$ by $\sin\phi=L_1/Q$, so $\cos\phi=W_1/Q$. Then use the identity $\sin\phi\cos\theta+\cos\phi\sin\theta=\sin(\phi+\theta)$.

Gerry Myerson
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If you write $\cos \theta = \sqrt {1-\sin^2 \theta}$, isolate that term and square, you get a quadratic in $\sin \theta$

Ross Millikan
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