To find the angles of the triangle with the vertices $P(2, 0)$, $Q(0, 3)$, and $R(3, 4)$, we can follow these steps:

Step 1: Find the lengths of the sides

We use the distance formula to find the lengths of the sides of the triangle. The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Side $PQ$ (between points $P(2, 0)$ and $Q(0, 3)$):

$PQ = \sqrt{(0 - 2)^2 + (3 - 0)^2} = \sqrt{(-2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61$

Side $PR$ (between points $P(2, 0)$ and $R(3, 4)$):

$PR = \sqrt{(3 - 2)^2 + (4 - 0)^2} = \sqrt{(1)^2 + (4)^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.12$

Side $QR$ (between points $Q(0, 3)$ and $R(3, 4)$):

$QR = \sqrt{(3 - 0)^2 + (4 - 3)^2} = \sqrt{(3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10} \approx 3.16$

Step 2: Use the Law of Cosines to find the angles

The Law of Cosines states: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ Where $a$, $b$, and $c$ are the lengths of the sides, and $C$ is the angle opposite side $c$.

We will use this formula to find each of the angles.

Angle $\angle PQR$ (opposite side $PR$):

$\cos \angle PQR = \frac{PQ^2 + QR^2 - PR^2}{2 \cdot PQ \cdot QR}$ Substituting the values: [ \cos \angle PQR = \frac{(3.61)^2 + (3.16)^2 - (4.12)^2}{2 \cdot 3.61 \cdot 3.16} = \frac{13.0321 + 9.9856 - 16.9744}{