Find the unknown forces and angle in the diagram drawn below. Diagram not drawn to scale.

The sum of angles around a point is $360^\circ$.
The given angles are $150^\circ$ (between P and 150 N) and $120^\circ$ (between 150 N and Q).

$$\text{Angle between P and Q} = 360^\circ - 150^\circ - 120^\circ$$ $$\text{Angle between P and Q} = 360^\circ - 270^\circ$$ $$\text{Angle between P and Q} = \boxed{90^\circ}$$

Lami's Theorem states that if three coplanar, concurrent forces are in equilibrium, then each force is proportional to the sine of the angle between the other two forces.
Let $F_1 = P$, $F_2 = Q$, and $F_3 = 150\, \text{N}$.
The angle opposite to P is $120^\circ$.
The angle opposite to Q is $150^\circ$.
The angle opposite to 150 N is $90^\circ$.

$$\frac{P}{\sin 120^\circ} = \frac{Q}{\sin 150^\circ} = \frac{150\, \text{N}}{\sin 90^\circ}$$

Using Lami's Theorem for P:

$$\frac{P}{\sin 120^\circ} = \frac{150}{\sin 90^\circ}$$

We know that $\sin 120^\circ = \sin (180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\sin 90^\circ = 1$.

$$P = 150 \times \frac{\sin 120^\circ}{\sin 90^\circ}$$ $$P = 150 \times \frac{\frac{\sqrt{3}}{2}}{1}$$ $$P = 75\sqrt{3}\, \text{N}$$ $$P \approx 75 \times 1.732$$ $$P \approx \boxed{129.9\, \text{N}}$$

Using Lami's Theorem for Q:

$$\frac{Q}{\sin 150^\circ} = \frac{150}{\sin 90^\circ}$$

We know that $\sin 150^\circ = \sin (180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$ and $\sin 90^\circ = 1$.

$$Q = 150 \times \frac{\sin 150^\circ}{\sin 90^\circ}$$ $$Q = 150 \times \frac{\frac{1}{2}}{1}$$ $$Q = \boxed{75\, \text{N}}$$

The unknown forces are $P \approx \textbf{129.9 N}$ and $Q = \textbf{75 N}$. The unknown angle between P and Q is $\textbf{90°}$.