# 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 $A=\sqrt{B^2 + B^2 sen^2(a)}$.