My problem:
I have: $x$, $y$ & $\alpha$ and the aspect ratio $o$(long):$p$(short) (red rectangle)
I want to have $n$ & $m$ in dependancy of $x, y, \alpha, o, p$
I tried to figure it out with cos, sin and tan but I don't get a solution. My math teacher said something about the matrix of rotation, but I don't know this method.
I'm also fine with the points where the red corners are on the black rectangle.
Is this possible ?
Thanks in advance ;)
Let $z$ be the length of the hypotenuse of the top left right-angled triangle whose two sides are $m$ and $n$. Then $z$ is the length of one side of the red rectangle. Let the length of the other side be $kz$. If we don't know which side is longer in advance, we don't know whether $k$ is equal to $\frac po$ or $\frac op$. However, we do have $$ \begin{cases} z\sin\alpha + kz\cos\alpha = x,\tag{1}\\ z\cos\alpha + kz\sin\alpha = y. \end{cases} $$ Hence we can determine $k$ without using $o$ or $p$: $$ k = \frac{y\sin\alpha - x\cos\alpha}{x\sin\alpha - y\cos\alpha}.\tag{2} $$ Now, from $(1)$, we get $$ z = \frac x{\sin\alpha + k\cos\alpha}. $$ Therefore $$ \begin{align} n &= z\sin\alpha = \frac{x\sin\alpha}{\sin\alpha + k\cos\alpha},\\ m &= z\cos\alpha = \frac{x\cos\alpha}{\sin\alpha + k\cos\alpha}. \end{align} $$