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 three angles of a triangle with vertices \( P(2, 0) \), \( Q(0, 3) \), and \( R(3, 4) \), we can follow these steps:

### Step 1: Calculate the lengths of the sides
We can use the distance formula to find the lengths of the sides of the triangle:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

#### 1. Length of side \( PQ \) (between points \( P(2, 0) \) and \( Q(0, 3) \)):
\[
PQ = \sqrt{(0 - 2)^2 + (3 - 0)^2} = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}
\]

#### 2. Length of side \( PR \) (between points \( P(2, 0) \) and \( R(3, 4) \)):
\[
PR = \sqrt{(3 - 2)^2 + (4 - 0)^2} = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}
\]

#### 3. Length of side \( QR \) (between points \( Q(0, 3) \) and \( R(3, 4) \)):
\[
QR = \sqrt{(3 - 0)^2 + (4 - 3)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}
\]

So, we have the side lengths:
- \( PQ = \sqrt{13} \)
- \( PR = \sqrt{17} \)
- \( QR = \sqrt{10} \)

### Step 2: Apply the Law of Cosines to find the angles

The Law of Cosines is given by:
\[
\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}
\]
where \( C \) is the angle opposite side \( c \), and \( a \), \( b \), and \( c \) are the lengths of the sides.

#### 1. Find angle \( \angle P \) (opposite side \( QR \)):
\[
\cos(\angle P) = \frac{PQ^2 + PR^2 - QR^2}{2 \times PQ \times PR}
\]
Substituting the values:
\[
\cos(\angle P) = \frac{(\sqrt{13})^2 + (\sqrt{17})^2 - (\sqrt{10})^2}{2 \times \sqrt{13} \times \sqrt{17}} = \frac{13 + 17 - 10}{2 \times \sqrt{13} \times \sqrt{17}} = \frac{20}{2 \times \sqrt{13} \times \sqrt{17}}
\]
Calculating this:

#### 2. Find angle \( \angle Q \) (opposite side \( PR \)):
\[
\cos(\angle Q) = \frac{PQ^2 + QR^2 - PR^2}{2 \times PQ \times QR}
\]

#### 3. Find angle \( \angle R \) (opposite side \( PQ \)):
\[
\cos(\angle R) = \frac{PR^2 + QR^2 - PQ^2}{2 \times PR \times QR}
\]

Let me compute the exact values for the angles now.The three angles of the triangle with vertices \( P(2, 0) \), \( Q(0, 3) \), and \( R(3, 4) \), correct to the nearest degree, are:

- \( \angle P = 48^\circ \)
- \( \angle Q = 75^\circ \)
- \( \angle R = 58^\circ \)

Would you like further details or explanations on any part of the process?

Here are 5 related questions for further exploration:
1. How can the Law of Cosines be derived from the Pythagorean theorem?
2. How does the Law of Sines relate to the angles and sides of a triangle?
3. What are the properties of a triangle that make it obtuse, acute, or right-angled?
4. Can we solve the same problem using vectors? If so, how?
5. How would the problem change if one of the points was at the origin?

**Tip:** The sum of all interior angles of a triangle is always \(180^\circ\), a helpful check for your calculations!

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) using the Law of Cosines. This problem demonstrates calculating side lengths using the distance formula and applying the Law of Cosines to find angles to the nearest degree. Suitable for students in Grades 9-11.

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