Step 1: Redraw the circuit for open circuit conditions. Pull the load resistor $R_{Load}$ out of the circuit. The two terminals left, node B and the ground rail, are where we measure the open circuit (Thevenin) voltage. Node B connects to the rest of the circuit only through the 100 $\Omega$ resistor $R_3$. With the load gone, node B is a dead end, so $R_3$ passes zero current.
Step 2: Simplify the source side. Because $R_3$ carries no current, we can ignore it completely while finding the open circuit voltage, and the problem reduces to a plain two resistor voltage divider made of the source (15 V), $R_1 = 100\ \Omega$, and $R_2 = 100\ \Omega$. This divider carries a loop current of \[ I = \frac{15}{100+100} = 0.075\ \text{A} \]
Step 3: Read off the divider voltage. The voltage across $R_2$, which is node A with respect to ground, is \[ V_A = 0.075 \times 100 = 7.5\ \text{V} \] Since $R_3$ has zero drop, the voltage at node B (same as at the open $R_{Load}$ terminals) is also 7.5 V. This is $V_{th}$.
Step 4: Cross check with Thevenin resistance. As a sanity check, find $R_{th}$ by shorting the source and looking into the terminals from node B. Looking back from node B, $R_3$ (100 $\Omega$) is in series with $R_1 \parallel R_2$ (both 100 $\Omega$). \[ R_1 \parallel R_2 = \frac{100 \times 100}{100+100} = 50\ \Omega, \quad R_{th} = 100 + 50 = 150\ \Omega \] This does not change $V_{th}$, it only confirms the Thevenin equivalent is consistent, and $V_{th}$ stays at the divider result found above.
Final check: Matching against the choices, 7.5 V is option (A), while 5.0, 15.0 and 10.0 all come from miscounting which resistors actually share current with the load branch. \[ \boxed{V_{th} = 7.5\ \text{V}} \]
