How to find vertices of a rectangle when center coordinates and angle of tilt is given? Need to find z1, z2 z3 and z4 when zc and r are given.
Been trying to solve this seemingly simple geometry problem. Can't wrap my head around it.
It's a rectangle with center coordinates given and the angle of tilt. We also have the length of each sides (2L and 2B). We need to find the vertices.
 A: 
If $\gamma = 0$, vector $\vec{u} = ( w, 0 )$ and vector $\vec{v} = ( 0, b )$.
Rotating a 2D vector $(x , y)$ counterclockwise yields
$$\begin{cases} x^, = x \cos\gamma - y \sin\gamma \\ y^, = x \sin\gamma + y \cos\gamma \end{cases} \tag{1}\label{na1}$$
therefore
$$\vec{u} = ( w \cos\gamma ,\, w \sin\gamma ) \tag{2}\label{na2}$$
and
$$\vec{v} = ( -b \sin\gamma ,\, b \cos\gamma ) \tag{3}\label{na3}$$
If we know $\gamma$, $w$, $b$, and $\vec{z_c}$, then the four vertices of the rotated rectangle are
$$\begin{cases}
\vec{z_1} = \vec{z_c} - \vec{u} + \vec{v} \\
\vec{z_2} = \vec{z_c} + \vec{u} + \vec{v} \\
\vec{z_3} = \vec{z_c} + \vec{u} - \vec{v} \\
\vec{z_4} = \vec{z_c} - \vec{u} - \vec{v} \end{cases} \tag{4}\label{na4}$$
or equivalently,
$$\begin{cases}
x_1 = x_c - w \cos\gamma - b \sin\gamma \\
x_2 = x_c + w \cos\gamma - b \sin\gamma \\
x_3 = x_c + w \cos\gamma + b \sin\gamma \\
x_4 = x_c - w \cos\gamma + b \sin\gamma \end{cases}, \qquad \begin{cases}
y_1 = y_c - w \sin\gamma + b \cos\gamma \\
y_2 = y_c + w \sin\gamma + b \cos\gamma \\
y_3 = y_c + w \sin\gamma - b \cos\gamma \\
y_4 = y_c - w \sin\gamma - b \cos\gamma \end{cases} \tag{5}\label{na5}$$
A: Hint:
Consider the coordinate system with $z_c=(0,0)$:

