1. Solve the trigonometric equation: $5\sin\phi + 3 = 0$ for $0^\circ \leq \phi \leq 360^\circ$. 2. Rearrange the equation to isolate $\sin\phi$: $$5\sin\phi = -3$$ $$\sin\phi = -\frac{3}{5} = -0.6$$ 3. Recall that $\sin\phi$ is negative in the third and fourth quadrants. 4. Find the reference angle $\alpha$ where $\sin\alpha = 0.6$: $$\alpha = \sin^{-1}(0.6) \approx 36.87^\circ$$ 5. Calculate the solutions in the specified range: - Third quadrant: $\phi = 180^\circ + \alpha = 180^\circ + 36.87^\circ = 216.87^\circ$ - Fourth quadrant: $\phi = 360^\circ - \alpha = 360^\circ - 36.87^\circ = 323.13^\circ$ 6. Final solutions: $$\phi = 216.87^\circ, 323.13^\circ$$ --- Note: Part 1 solved as requested. Part 2 involves geometric construction and measurement, which cannot be performed here.