P(3,2), Q(-1,5) and R(2,1) are vertices of a triangle. 1) Find the length of the sides of the triangle. 11) Use the dot product to find an angle of the triangle.

• Length of PQ: For P$(3, 2)$ and Q$(-1, 5)$ $$PQ = \sqrt{(-1 - 3)^2 + (5 - 2)^2} = \sqrt{(-4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$
• Length of QR: For Q$(-1, 5)$ and R$(2, 1)$ $$QR = \sqrt{(2 - (-1))^2 + (1 - 5)^2} = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
• Length of RP: For R$(2, 1)$ and P$(3, 2)$ $$RP = \sqrt{(3 - 2)^2 + (2 - 1)^2} = \sqrt{(1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$$

The lengths of the sides are:

• $PQ = \boxed{5}$
• $QR = \boxed{5}$
• $RP = \boxed{\sqrt{2}}$

Let's find the angle at vertex P, $\angle QPR$. First, determine the vectors $\vec{PQ}$ and $\vec{PR}$.

$$\vec{PQ} = Q - P = (-1 - 3, 5 - 2) = \begin{bmatrix} -4 \\ 3 \end{bmatrix}$$ $$\vec{PR} = R - P = (2 - 3, 1 - 2) = \begin{bmatrix} -1 \\ -1 \end{bmatrix}$$

$$|\vec{PQ}| = \sqrt{(-4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$ $$|\vec{PR}| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$$

$$\cos(\angle QPR) = \frac{\vec{PQ} \cdot \vec{PR}}{|\vec{PQ}| |\vec{PR}|} = \frac{1}{5 \cdot \sqrt{2}} = \frac{1}{5\sqrt{2}}$$

Rationalize the denominator:

$$\cos(\angle QPR) = \frac{1}{5\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{10}$$

$$\angle QPR = \arccos\left(\frac{\sqrt{2}}{10}\right) \approx \arccos(0.14142) \approx 81.87^\circ$$

An angle of the triangle is $\boxed{\arccos\left(\frac{\sqrt{2}}{10}\right) \approx 81.87^\circ}$.