Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In web graphics if you ask for the width property of a rectangle you'll get back the horizontal measurement of the rectangle at that time rather than it's starting width. So a rectangle that is longer than tall will return a shorter width measurement when it is rotated.

I'd like to measure it's unrotated width.

I thought I had a solution with the code below, however it works for all angles except at 45, -45, 135, and -135.

public static function width( designObject:DisplayObject ):Number
    // convert degrees to radians
    var r:Number = designObject.rotation * Math.PI/180;
    // cos, c in the equation
    var c:Number = Math.abs(Math.cos(r));
    // sin, s in the equation
    var s:Number = Math.abs(Math.sin(r));
    // get the unrotated width
    var w:Number = (designObject.width * c - designObject.height * s) / (Math.pow(c, 2) - Math.pow(s, 2));

    return w;

Using the code above a rectangle that is 148.2 pixels wide returns that width for all angles, but at 45 degrees it returns 64. Am I missing something, might there be an alternate solution?


share|improve this question

1 Answer 1

It's probably bugging out because $\cos 45^\circ = \sin 45^\circ$, so the denominator becomes zero. I can't imagine why it would return exactly $64$ though.

In any case, there probably isn't any way to do it for exactly $45^\circ$. For any given square, there are lots of rectangles of different sizes that fit exactly in it at $45^\circ$. So just from knowing the size of the square, you can't recover the dimensions of the rectangle. You can do it if you know the aspect ratio of the rectangle, though.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.