In triangle $ABC$, $\angle C = 90^\circ$, $\angle A = 40^\circ$, and $AC = 10$. Find the radius of the incircle of triangle $ABC$.
Let $I$ be the incenter of triangle $ABC$, and let $M$ be the midpoint of $\overline{AC}$, as shown below.
[asy]
unitsize(1 cm);
pair A, B, C, I, M;
C = (0,0);
A = (8,0);
B = intersectionpoint(arc(C,10,0,180),arc(A,24,0,180));
I = incenter(A,B,C);
M = (A + C)/2;
draw(A--B--C--cycle);
draw(A--I--C);
draw(C--M);
label("$A$", A, ENE);
label("$B$", B, NW);
label("$C$", C, SW);
label("$I$", I, S);
label("$M$", M, S);
[/asy]
We draw the incircle of triangle $ABC$ and locate its center $I$.
[asy]
unitsize(1 cm);
filter grayscale;
pair A, B, C, I, M;
C = (0,0);
A = (8,0);
B = intersectionpoint(arc(C,10,0,180),arc(A,24,0,180));
I = incenter(A,B,C);
M = (A + C)/2;
draw(A--B--C--cycle);
draw(incircle(A,B,C));
draw(A--I--C);
draw(C--M);
label("$A$", A, ENE);
label("$B$", B, NW);
label("$C$", C, SW);
label("$I$", I, SE);
label("$M$", M, S);
label("$10/2 = 5$", (A + M)/2, S);
[/asy]
Let $r$ be the inradius of triangle $ABC$, and let $E$ be one of the points of tangency of the incircle of triangle $ABC$ with $\overline{AC}$. Then $IE = r$, so by the Pythagorean Theorem on triangle $IMC$, we have $r^2 + (5 - r)^2 = I^2$. Since $I$ lies on the angle bisector of $\angle C$, we have
\[\frac{AC}{IC} = \frac{AE}{EI}.\]In other words,
\[\frac{10}{IC} = \frac{5}{r}.\]Then $\frac{IC}{5} = \frac{r}{10}$. Since $\angle ABC = 50^\circ$, $\angle IBC = 90^\circ - 50^\circ = 40^\circ$, so $\angle ICB = 180^\circ - 90^\circ - 40^\circ = 50^\circ$. Therefore, triangle $IBC$ is isosceles, and $IC = IB$.
Hence, $\frac{IC + IB}{5} = \frac{2 IC}{5} = \frac{r}{10}$, so $2 IC r = 5 \cdot 10 = 50$. But $r^2 + (5 - r)^2 = I^2 = 50$. Hence,
\begin{align*}
2 IC r &= 2 IC \cdot 5 \cdot \sqrt{r^2 + (5 - r)^2} \\
&= 2 IC \cdot 5 \cdot \sqrt{50} \\
&= 2 \cdot 5 IC \cdot 5 \\
&= 100 IC,
\end{align*}so $100 IC = 50$, and $IC = \frac{1}{2}$. Therefore, $r = IC \cdot \frac{r}{10} = \boxed{\frac{1}{20}}$.
To find the radius of the incircle of triangle $ABC$, we can make use of the fact that the inradius is given by the formula:
$$r = \frac{A}{s},$$
where $A$ is the area of the triangle and $s$ is the semiperimeter (half of the perimeter).
Since we're given that $\angle C = 90^\circ$, we can use the Pythagorean theorem to find the length of $BC$:
$$BC^2 = AB^2 + AC^2.$$
Since $\angle A = 40^\circ$, we know that $\angle B = 180^\circ - \angle A - \angle C = 180^\circ - 40^\circ - 90^\circ = 50^\circ$.
Let's denote $BC$ as $x$. Thus, we have:
$$x^2 = AB^2 + AC^2.$$
From the given information, we know that $AC = 10$, so let's replace that in the equation:
$$x^2 = AB^2 + 10^2.$$
Since we have two angles ($40^\circ$ and $50^\circ$), we can use the trigonometric identity $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ to find the length of $AB$ in terms of $x$, which is adjacent to the angle $\angle B$:
$$\tan(40^\circ) = \frac{AB}{BC}.$$
Rearranging this equation, we get:
$$AB = BC \cdot \tan(40^\circ).$$
Substituting this into the earlier equation:
$$x^2 = (BC \cdot \tan(40^\circ))^2 + 10^2.$$
Simplifying further, we have:
$$x^2 = B^2 + 100.$$
Recall that the semiperimeter $s$ is given by:
$$s = \frac{AB + BC + AC}{2}.$$
Now, we can substitute our expressions for $AB$ and $AC$ into the equation for the semiperimeter:
$$s = \frac{BC \cdot \tan(40^\circ) + BC + 10}{2}.$$
To find the area of the triangle, we can use Heron's Formula:
$$A = \sqrt{s(s - AB)(s - BC)(s - AC)}.$$
Substituting the expressions for $AB$, $BC$, and $AC$ into the equation for the area, we get:
$$A = \sqrt{s \left(s - BC \cdot \tan(40^\circ)\right)\left(s - BC\right)\left(s - 10\right)}.$$
Finally, plugging the expressions for the area $A$ and semiperimeter $s$ into the formula for the inradius $r$, we obtain:
$$r = \frac{\sqrt{s \left(s - BC \cdot \tan(40^\circ)\right)\left(s - BC\right)\left(s - 10\right)}}{s}.$$