# How to calculate the new intersection on the x-axis after rotation of a rectangle?

I've been trying to calculate the new intersection on the x-axis after rotation of any given rectangle. The rectangle's center is the point $(0,0)$.

What do I know:

• length of B (that is half of the width of the given rectangle)
• angle of a (that is the rotation of the rectangle)

What do I want to know: length of A (or value of point c on the x-axis).

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By the Law of Sines and since $b$ is a right angle, $$len(A) = \frac{len(B)}{sin(\frac{\pi}{2}-a)}$$ where $0 \leq a <\pi$.

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@user6312 Oh, I was just going by his diagram – Nicolas Villanueva May 2 '11 at 15:10
The width of the rectangle is indeed 2B. My wording was wrong in the original question. – Webdevotion May 2 '11 at 15:13
b is not a right angle in my opinion. – Webdevotion May 3 '11 at 12:36
@Webdevotion b should be a right angle, else it's not a rectangle. Remember, we are given a s.t. B is perpendicular to C. – Nicolas Villanueva May 3 '11 at 15:16

Hint: Try to divide the cases. Referring to your image, after the rotation of the angle $a$ the vertex on the left side of the rectangle pass or not pass the x-axis?

Suppose now that your rectangle has one side of lenght 2B, and the other one "large", so the vertex on the left side doesn't pass the x-axis. Then using Pythagoras you get $A=\sqrt{B^2 + B^2 sen^2(a)}$.

What about the other case?

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I've tried your formula, but did not get the expected result. I also had a little bit of a problem understanding your comment. Could you rephrase please? "Suppose now that your rectangle has one side of lenght 2B, and the other one 'large', so the vertex on the left side doesn't pass the x-axis." – Webdevotion May 3 '11 at 12:43