I have drawn an image that should help shed some light on how to calculate what you desire. Drawing a simple picture is usually the best way to solve geometry questions.
$\begin{align} \tan x&=\frac{\text{opp}}{\text{adj}}=\frac{15}4=3.75\\ x&=\arctan3.75\approx75° \end{align}$

I am assuming by angle of the cone that you were alluding to the angle that the truncated cone makes with it's base, and that is what I have shown below.
It is clear from the diagram we have a triangle with sides of length $15\ \mathrm{cm}$ and $4\ \mathrm{cm}$, using the knowledge that $tan(x) = \frac{\text{opposite}}{\text{adjacent}}$ we have $tan(x) = \frac{15}{4} = 3.75$. Which when taking the inverse tangent we find that $x = 75.06858°$.
More information on trigonometry can be found at http://en.wikipedia.org/wiki/Trigonometry
Based on the revised question asked the angle you are interested in can be calculated as follows:
First we need to calculate the height of the non-truncated code. This is calculated as the height of the truncated cone multiplied by the ratio of the radius base of the cone and the difference in radius of the base and the top of the truncated cone.
$t = h \times \frac{b}{b-a} = 15 \times \frac{24}{4} = 90$
Here $t$ is total height of the cone, $h$ is height of the truncated cone, $b$ is radius of the base and a is radius of the top of the truncated cone.
We can now use pythag to find the length of the outside of the full height cone.
$r = \sqrt{t^2 + b^2} = 93.145$
The arc length of the flattened out cone is the circumference of the of base of the cone which is $\pi d$ (or $180d$ in degrees). So our arc length $l = 180 \times 2 \times 24 = 8640$
We know that arc length is $\text{radius} \times \text{angle}$ so:
$8640 = l = xr = x 93.145$
rearranging gives $x = 92.9786$