1. **State the problem:** We need to find the angle $\theta$ such that $0^\circ \leq \theta \leq 90^\circ$ and satisfies the equation $$3 \sin\left(\frac{\theta}{2} - 20^\circ\right) = 0.85.$$\n\n2. **Isolate the sine function:** Divide both sides by 3 to get \n$$\sin\left(\frac{\theta}{2} - 20^\circ\right) = \frac{0.85}{3} = 0.2833.$$\n\n3. **Use the inverse sine function:** To solve for the angle inside the sine, use \n$$\frac{\theta}{2} - 20^\circ = \sin^{-1}(0.2833).$$\n\n4. **Calculate the principal value:** \n$$\sin^{-1}(0.2833) \approx 16.45^\circ.$$\n\n5. **Consider the sine function's periodicity:** Since sine is positive in the first and second quadrants, the second solution is \n$$180^\circ - 16.45^\circ = 163.55^\circ.$$\n\n6. **Solve for $\theta$ for both cases:**\n- Case 1: $$\frac{\theta}{2} - 20^\circ = 16.45^\circ \implies \frac{\theta}{2} = 36.45^\circ \implies \theta = 72.9^\circ.$$\n- Case 2: $$\frac{\theta}{2} - 20^\circ = 163.55^\circ \implies \frac{\theta}{2} = 183.55^\circ \implies \theta = 367.1^\circ.$$\n\n7. **Check the domain:** Since $\theta$ must be between $0^\circ$ and $90^\circ$, only $\theta = 72.9^\circ$ is valid.\n\n**Final answer:** $$\boxed{\theta = 72.9^\circ}.$$