The radius of the circumscribed and inscribed circle of a triangle is 170 and 12 units, respectively, and the perimeter of the triangle is 416 units. What are the angles of the triangle?
1. Let $R$ and $r$ denote the given radii and $2s$ the perimeter. We can consider the data set symmetric in the sense that no side or angle is distinguished from the others. Therefore, if we somehow obtain an equation for one side, it must also hold for the other two sides, implying that it must be at least a cubic equation. In the absence of other prospects, we seek such an equation.

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The known relationships are $4Rt = abc$ and $t = rs$, where $t$ is the area. Both are symmetric in $a, b, c$, and the Heron's formula for the area is also symmetric: $t^2 = s(s-a)(s-b)(s-c)$. These three relationships are sufficient to write down the cubic equation whose roots are $a, b, c$. Knowing these, we will then proceed to calculate the angles.

We expand the factored form of the equation:

$$
(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab + bc + ca)x - abc = 0
$$

Here, we already know the first and third coefficients: $abc = 4Rt = 4Rrs$. The second coefficient can be obtained from the expansion of Heron's formula:

$$
t^2 = r^2 s^2 = s\left(s^3 - (a+b+c)s^2 + (ab + bc + ca)s - abc\right)
$$

from which we get

$$
ab + bc + ca = \frac{1}{s}\left(r^2 s - s^3 + 2s^3 + abc\right) = r^2 + s^2 + 4Rr
$$

Thus, the equation is:

$$
x^3 - 2sx^2 + (r^2 + s^2 + 4Rr)x - 4Rrs = 0
$$

With our data, this becomes:

$$
x^3 - 416x^2 + 51568x - 1697280 = 0
$$

2. Let's try to find if the equation has an integer root. This can only be a divisor of the constant term, since the first three terms are divisible by $x$. Factoring the constant term into prime powers, we get $1697280 = 2^9 \cdot 3 \cdot 5 \cdot 13 \cdot 17$. Let's try the divisors $2^5 = 32$ and $2^6 = 64$: Denote the left-hand side of the equation by $P(x)$, so $P(32) = -4403200$. These are of opposite signs, and since $P(x)$ is a continuous function, it must have a root in the interval $(32, 64)$. Based on the magnitude of the errors, $-440$ thousand and $+161$ thousand, the root is expected to be closer to 64. The divisors of the constant term are $3 \cdot 17 = 51$ and $2^2 \cdot 13 = 52$. Indeed, $P(52) = 0$, so $x_1 = 52$.

The sum of the other two roots is: $x_2 + x_3 = x_1 + x_2 + x_3 - 52 = 416 - 52 = 364$, and their product is: $x_2 x_3 = x_1 x_2 x_3 / x_1 = x_1 x_2 x_3 / (2^2 \cdot 13) = 2^7 \cdot 3 \cdot 5 \cdot 17 = 32640$, so the other two sides are the roots of the following quadratic equation:

$$
x^2 - 364x + 32640 = 0, \quad x_2 = 160, \quad x_3 = 204
$$

Thus, the sides of the triangle are $a = 52, b = 160, c = 204$. In this triangle, $R, r$, and the perimeter are indeed equal to the given measurements.

3. We calculate the angles opposite the two smaller sides, which are definitely acute angles:

$$
\begin{gathered}
\sin \alpha = \frac{a}{2R} = \frac{52}{340}, \quad \sin \beta = \frac{b}{2R} = \frac{160}{340} \\
\alpha = 8^\circ 47.8', \quad \beta = 28^\circ 4.4' \quad \text{and finally} \quad \gamma = 143^\circ 7.8'
\end{gathered}
$$

Remark. We can work with smaller numbers if we write a cubic equation for the segments $s-a, s-b, s-c$ of the incircle. These roots have symmetric expressions:

$$
\begin{gathered}
(s-a) + (s-b) + (s-c) = s \\
(s-a)(s-b) + (s-b)(s-c) + (s-a)(s-c) = 3s^2 - s \cdot 4s + (ab + ac + bc) = r^2 + 4Rr \\
(s-a)(s-b)(s-c) = t^2 / s = r^2 s
\end{gathered}
$$

Thus, these are the roots of the following equation:

$$
z^3 - sz^2 + (r^2 + 4Rr)z - r^2 s = 0
$$

With our data, this becomes:

$$
z^3 - 208z^2 + 8304z - 29952 = 0
$$

Since $2s$ is an even number, we can equally hope that $s-a, \ldots$ are integers, as in the previous attempt. From here, with $z_1 = 156, z_2 = 48, z_3 = 4$, we get $a = 208 - 156 = 52$ and proceed to the previous results.