Let's assume that you knew what you were doing when you wrote:
I tried finding the Thevenin equivalent of the input loop by using voltage divider to find Vb and Rb. And using Kirchoff's Voltage law on the input loop.
I'm willing to accept your assurances because you wrote that close enough to what I might expect. So let's just write that out (I'll assume here that you know why I'm writing it out this way without having to re-draw the schematic and point out the obvious to you):
$$\begin{align*}R_\text{TH}&=\frac{R_1\cdot R_2}{R_1+R_2}&V_\text{TH}&=V_\text{CC}\cdot\frac{R_1}{R_1+R_2}\\\\I_\text{B}&=\frac{V_\text{TH}-V_\text{BE}}{R_\text{TH}+\left(\beta+1\right)\cdot R_\text{E}}&I_\text{C}&=\beta\cdot I_\text{B} & I_\text{E}&=\left(\beta+1\right)\cdot I_\text{B}\\\\V_\text{B}&=V_\text{TH}-I_\text{B}\cdot R_\text{TH}&V_\text{C}&=V_\text{CC}-I_\text{C}\cdot R_\text{C}&V_\text{E}&=I_\text{E}\cdot R_\text{E} \end{align*}$$
From this, you can easily compute:
$$\begin{align*}V_\text{CE}&=V_\text{C}-V_\text{E}=V_\text{CC}-I_\text{C}\cdot R_\text{C}-I_\text{E}\cdot R_\text{E}\\\\&=V_\text{CC}-\beta\cdot I_\text{B}\cdot R_\text{C}-\left(\beta+1\right)\cdot I_\text{B}\cdot R_\text{E}\\\\ &=V_\text{CC}-I_\text{B}\cdot \big[\beta\cdot R_\text{C}-\left(\beta+1\right)\cdot R_\text{E}\big]\\\\&=V_\text{CC}-\frac{V_\text{TH}-V_\text{BE}}{R_\text{TH}+\left(\beta+1\right)\cdot R_\text{E}}\cdot \big[\beta\cdot R_\text{C}-\left(\beta+1\right)\cdot R_\text{E}\big] \end{align*}$$
Plot that as a function of \$\beta\$ and see what you get. What is the value with \$\beta=0\$? What is the value as \$\beta\to\infty\$? I think you should now be able to see why a precise value for \$\beta\$ isn't necessary here.