Find the values of θ between 0° and 360° satisfying the equation 5 sin θ = -4.

Step 1: The volume $V$ of a right circular cone is given by the formula
$V = \frac{1}{3} \pi r^{2} h\ cm^{3},$
where $r$ is the radius and $h$ is the height.

Given the constraint
$r + h = 5\ cm.$

Solve for $h$:
$h = 5 - r\ cm.$

Step 2: Substitute $h = 5 - r$ into the volume formula:
$V = \frac{1}{3} \pi r^{2} (5 - r)\ cm^{3}.$

Expand:
$V = \frac{1}{3} \pi (5 r^{2} - r^{3})\ cm^{3}.$

Step 3: To find the maximum volume, compute the derivative $\frac{dV}{dr}$:
$\frac{dV}{dr} = \frac{1}{3} \pi \frac{d}{dr} (5 r^{2} - r^{3}).$

First, find the derivative of the inside:
$\frac{d}{dr} (5 r^{2} - r^{3}) = 10 r - 3 r^{2}.$

So,
$\frac{dV}{dr} = \frac{1}{3} \pi (10 r - 3 r^{2})\ cm^{3/cm}.$

Step 4: Set the derivative equal to zero to find critical points:
$\frac{1}{3} \pi (10 r - 3 r^{2}) = 0.$

Since $\frac{1}{3} \pi \neq 0$, solve
$10 r - 3 r^{2} = 0.$

Factor:
$r (10 - 3 r) = 0.$

Thus,
$r = 0\ cm\quad or\quad r = \frac{10}{3}\ cm.$

Step 5: Check the endpoints and critical points. Domain: $0 < r < 5\ \mathrm{cm}$ (since $h > 0$).

At $r \to 0^{+}$, $V \to 0\ cm^{3}$.
At $r \to 5^{-}$, $V \to 0\ cm^{3}$.

At $r = \frac{10}{3}\ \mathrm{cm}$, this is a candidate for maximum.

Corresponding $h$:
$h = 5 - \frac{10}{3} = \frac{15}{3} - \frac{10}{3} = \frac{5}{3}\ cm.$

Step 6: Compute the second derivative to confirm maximum:
$\frac{d^{2}V}{dr^{2}} = \frac{1}{3} \pi \frac{d}{dr} (10 r - 3 r^{2}) = \frac{1}{3} \pi (10 - 6 r).$

At $r = \frac{10}{3}\ \mathrm{cm}$:
$10 - 6 \cdot \frac{10}{3} = 10 - 20 = -10 < 0,$
so it is a local maximum.

Step 7: Substitute $r = \frac{10}{3}\ \mathrm{cm}$ and $h = \frac{5}{3}\ \mathrm{cm}$ into the volume formula:
$V = \frac{1}{3} \pi \left( \frac{10}{3} \right)^{2} \left( \frac{5}{3} \right)\ cm^{3}.$

First, compute $\left( \frac{10}{3} \right)^{2}$:
$\left( \frac{10}{3} \right)^{2} = \frac{100}{9}.$

Then,
$V = \frac{1}{3} \pi \cdot \frac{100}{9} \cdot \frac{5}{3}\ cm^{3}.$

Multiply:
$\frac{100}{9} \cdot \frac{5}{3} = \frac{500}{27}.$

So,
$V = \frac{1}{3} \pi \cdot \frac{500}{27} = \frac{500 \pi}{81}\ cm^{3}.$

The maximum volume is $\boxed{\dfrac{500\pi}{81}\ cm^{3}}$.