Calculating for current in a single transistor amplifier(almost common collector?)
Just redrawing your schematic slightly and adding some labels:







simulate this circuit &ndash; Schematic created using CircuitLab

Just use KVL to start, following around through the base:

$$\begin{align*}
V - I_{Rc}\cdot R_c - I_{Rb}\cdot R_b - V_{BE} - I_E\cdot R_e &amp;= 0
\end{align*}$$

If the BJT is in its active region where your value of \$\beta=175\$ applies (and you'll know one way or another, soon enough), then it also follows that:

$$\begin{align*}
I_{Rc} &amp;= I_C+I_B=I_E \\
I_{Rb} &amp;= I_B \\
I_E&amp;=\left(\beta+1\right)\cdot I_B
\end{align*}$$

Applying those to the original equation, we get:

$$\begin{align*}
V - I_E\cdot R_c - I_B\cdot R_b - V_{BE} - I_E\cdot R_e &amp;= 0 \\
  \\
V - \left(\beta+1\right)\cdot I_B\cdot R_c - I_B\cdot R_b - V_{BE} - \left(\beta+1\right)\cdot I_B\cdot R_e &amp;= 0 \\
  \\
V &amp;= V_{BE} + I_B\cdot\left[\left(\beta+1\right)\cdot\left(R_c + R_e\right) + R_b \right] \\
  \\
I_B &amp;= \frac{V - V_{BE}}{R_b+\left(\beta+1\right)\cdot\left(R_c + R_e\right)}
\end{align*}$$

And that pretty much cracks the puzzle.

(It ignores little-re, which might have an impact in some cases but probably doesn't, here. It's impact is probably below 1%. But you could work it back into the equation later, if it matters to you.)

At this point, given your values and using \$V_{BE}=700\:\textrm{mV}\$, I get:

$$\begin{align*}
I_B&amp;\approx 10.1\:\mu\textrm{A} \\
I_E&amp;\approx 1.78\:\textrm{mA}
\end{align*}$$

So, I'd estimate:

$$\begin{align*}
V_E&amp;=I_e\cdot R_e\approx 1.78\:\textrm{V} \\
V_C&amp;=10\:\textrm{V}-I_E\cdot R_C\approx 6.44\:\textrm{V} \\
V_B&amp;=V_E+V_{BE}=V_C-I_B\cdot R_B\approx 2.48\:\textrm{V}
\end{align*}$$

Since \$V_{CE} \gt 1\:\textrm{V}\$, the BJT is in its active region and the value of \$\beta=175\$ can be considered to have applied, now that we can check it out. So it is fine to stop at this point and consider the question answered well enough.



Little re is because the base-emitter junction's thermal voltage can be treated as a tiny "battery" at the tip of the BJT emitter. It's always got the thermal voltage just sitting there, which at room temperature will be around \$26\:\textrm{mV}\$. Given a current through it (\$I_E\$), you can turn that into an equivalent resistance. This is called a lot of things, but in just talking I say "little re." In this case, \$re\approx 15\:\Omega\$.

This value adds to \$R_E\$ in the above calculations. Working it out, I find that it impacts the estimated current values by about 0.3%.

Just a note.