find correct to the nearest degree the three angles of the triangle with the given vertices. P(2,0), Q(0,3), R(3,4)

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}{

Find correct to the nearest degree the three angles of the triangle with the given vertices. P(2,0), Q(0,3), R(3,4)

Learn how to find the angles of a triangle with vertices P(2,0), Q(0,3), and R(3,4) by applying the distance formula and the Law of Cosines. This geometry problem covers key concepts such as triangle side lengths and calculating angles, making it suitable for students in Grades 9-12.

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