Formula to Calculate Each Pie Angle Where the Intercepted Arc is NOT the Center Point of the Circle AND All Slices are Equal Sized

Let’s place the point at which all of the cuts converge at the origin and the center of the circular pie at $(h,0)$ so that the circle can be parameterized as $x=h+r\cos t$, $y=r\sin t$. The parameter $t$ represents the angle at the center of the pie. If $\Gamma$ is the arc of the circle that goes from $t_1$ to $t_2$ then the area of the slice is $$\frac12\int_\Gamma x\,dy-y\,dx = \frac r2\int_{t_1}^{t_2}r+h\cos t\,dt = \frac r2\left(r(t_2-t_1)+h(\sin t_1-\sin t_0)\right).$$ If we want $n$ equally-sized slices, this area must be equal to $\pi r^2/n$, which leads to the equation $$rt_2+h\sin t_2 = \frac{2\pi r}n+rt_1+h\sin t_1.$$ If we fix $t_1$, this can be solved for $t_2$. Unfortunately, there’s no closed-form solution, but you can get a numerical approximation good enough for making the slices.

Taking your example, $h=\sqrt{1^2+1.5^2}\approx1.803$ and the area of each slice is approximately $10.603.$ The first cut is at $t=0$, and since there’s an even number of slices, we know that there will be another at $t=\pi$. By symmetry, we only need to compute two more cuts. Setting $t_1=0$ produces $t_2\approx 0.77$, and working backwards from the other cut, setting $t_2=\pi$ yields $t_1\approx 1.70$. The resulting pie slices look something like this:

If we relax the requirement that all of the cuts radiate from a common point, then there are many more ways to divvy up the pie.