Quick question about phase displacement in AC circuits
This can be solved using the data you have without any assumption, if (and this is not clear from what you said) the only thing you have to find is the active power provided by the voltage source, i.e. its average power.

Let's use phasors and let's assume all quantities are expressed using RMS values ("effective values"). As you may know the complex power \$S=P+jQ\$ (P and Q are the active and reactive power, respectively) absorbed by an impedance \$Z=R+jX\$ is given by:

$$
S = V \cdot I^*
$$

Where \$V\$ and \$I\$ are the voltage across \$Z\$ and the current through it, respectively. Using complex Ohm's law \$V=Z\cdot I\$ you can express S in two different ways:

\begin{align*}
S &amp;= Z \cdot |I|^2 = (R+jX) \cdot |I|^2 \\[1 em]
S &amp;= \dfrac {|V|^2} {Z^*} \; = \; \dfrac{R+jX}{R^2 + X^2} \cdot |V|^2
\end{align*}

From those formulas we can get the active power by considering only the real part of S:

\begin{align*}
P &amp;= Re(S) = R \cdot |I|^2 \\[1 em]
P &amp;= Re(S) = \dfrac{R}{R^2 + X^2} \cdot |V|^2
\end{align*}

We are interested in the active power \$P_s = Re(S_s)\$ provided by the source. We know from the conservation of energy, that it must be the sum of the active power absorbed by the two impedances:

$$
P_s = P_1 + P_2
$$

Since we know the RMS value of \$I_2\$, i.e. we know \$|I_2|\$, we can write:

$$
P_2 = R_2 \cdot |I_2|^2
\qquad\qquad
P_2 = \dfrac{R_2}{R_2^2 + X_2^2} \cdot |U|^2
$$

From those equations you can get \$P_2\$ and \$|U|^2\$. This latter allows to compute \$P_1\$ using the formula:

$$
P_1 = \dfrac{R_1}{R_1^2 + X_1^2} \cdot |U|^2
$$

As stated above, what you are looking for is \$P_s = P_1 + P_2\$.