The problem requires us to find the lengths of the sides AB, BC, and AC, and the measures of the angles $\angle ABC$, $\angle BAC$, and $\angle ACB$ for the triangle formed by the points A(3, 4), B(1, 1), and C(5, 1).

1. Problem Description

The problem requires us to find the lengths of the sides AB, BC, and AC, and the measures of the angles $\angle ABC$, $\angle BAC$, and $\angle ACB$ for the triangle formed by the points A(3, 4), B(1, 1), and C(5, 1).

2. Solution Steps

First, we calculate the lengths of the sides using the distance formula.

The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

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

$AB = \sqrt{(1-3)^2 + (1-4)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61$

$BC = \sqrt{(5-1)^2 + (1-1)^2} = \sqrt{(4)^2 + (0)^2} = \sqrt{16} = 4$

$AC = \sqrt{(5-3)^2 + (1-4)^2} = \sqrt{(2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61$

Next, we find the angles using the Law of Cosines.

The Law of Cosines is:

$c^2 = a^2 + b^2 - 2ab \cos(C)$, where a, b, and c are the side lengths, and C is the angle opposite side c.

To find $\angle ABC$, we use the formula with $AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos(\angle ABC)$:

$13 = 13 + 16 - 2(\sqrt{13})(4) \cos(\angle ABC)$

$0 = 16 - 8\sqrt{13} \cos(\angle ABC)$

$8\sqrt{13} \cos(\angle ABC) = 16$

$\cos(\angle ABC) = \frac{16}{8\sqrt{13}} = \frac{2}{\sqrt{13}}$

$\angle ABC = \cos^{-1}(\frac{2}{\sqrt{13}}) \approx 56.31^\circ$

To find $\angle BAC$, we use the formula with $BC^2 = AB^2 + AC^2 - 2(AB)(AC) \cos(\angle BAC)$:

$16 = 13 + 13 - 2(\sqrt{13})(\sqrt{13}) \cos(\angle BAC)$

$16 = 26 - 2(13) \cos(\angle BAC)$

$16 = 26 - 26 \cos(\angle BAC)$

$26 \cos(\angle BAC) = 10$

$\cos(\angle BAC) = \frac{10}{26} = \frac{5}{13}$

$\angle BAC = \cos^{-1}(\frac{5}{13}) \approx 67.38^\circ$

To find $\angle ACB$, we can use the fact that the sum of angles in a triangle is $180^\circ$:

$\angle ACB = 180^\circ - \angle ABC - \angle BAC = 180^\circ - 56.31^\circ - 67.38^\circ \approx 56.31^\circ$

3. Final Answer

AB = $\sqrt{13}$ ≈ 3.61

BC = 4

AC = $\sqrt{13}$ ≈ 3.61

m$\angle$ABC ≈ 56.31$^\circ$

m$\angle$BAC ≈ 67.38$^\circ$

m$\angle$ACB ≈ 56.31$^\circ$