We are given a triangle with vertices at $A(-5, -2)$, $B(4, -2)$, and $C(1, 3)$. To calculate the lengths of the sides and the angles of the triangle, we will follow these steps:

1. 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:

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

Side AB:

Using the coordinates $A(-5, -2)$ and $B(4, -2)$:

= \sqrt{(4 + 5)^2 + (0)^2} = \sqrt{9^2} = 9$$ #### **Side BC:** Using the coordinates $$B(4, -2)$$ and $$C(1, 3)$$: $$BC = \sqrt{(1 - 4)^2 + (3 - (-2))^2} = \sqrt{(-3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34}$$ #### **Side AC:** Using the coordinates $$A(-5, -2)$$ and $$C(1, 3)$$: $$AC = \sqrt{(1 - (-5))^2 + (3 - (-2))^2} = \sqrt{(1 + 5)^2 + (3 + 2)^2} = \sqrt{6^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61}$$ ### 2. Calculate the angles using the Law of Cosines. The Law of Cosines states: $$\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$$ Where $$a$$, $$b$$, and $$c$$ are the sides of the triangle and $$C$$ is the angle opposite side $$c$$. #### **Angle $$\angle ABC$$:** Using the sides $$AB = 9$$, $$BC = \sqrt{34}$$, and $$AC = \sqrt{61}$$: $$\cos(\angle ABC) = \frac{9^2 + \sqrt{34}^2 - \sqrt{61}^2}{2 \times 9 \times \sqrt{34}}$$ $$\cos(\angle ABC) = \frac{81 + 34 - 61}{2 \times 9 \times \sqrt{34}}$$ $$\cos(\angle ABC) = \frac{54}{18 \times \sqrt{34}} = \frac{3}{\sqrt{34}}$$ $$\angle ABC = \cos^{-1}\left(\frac{3}{\sqrt{34}}\right) \approx \cos^{-1}(0.514)$$ $$\angle ABC \approx 59.46^\circ$$ #### **Angle $$\angle BCA$$:** Using the sides $$AB = 9$$, $$BC = \sqrt{34}$$, and $$AC = \sqrt{61}$$: $$\cos(\angle BCA) = \frac{9^2 + \sqrt{61}^2 - \sqrt{34}^2}{2 \times 9 \times \sqrt{61}}$$ $$\cos(\angle BCA) = \frac{81 + 61 - 34}{2 \times 9 \times \sqrt{61}}$$ $$\cos(\angle BCA) = \frac{108}{18 \times \sqrt{61}} = \frac{6}{\sqrt{61}}$$ $$\angle BCA = \cos^{-1}\left(\frac{6}{\sqrt{61}}\right) \approx \cos^{-1}(0.770)$$ $$\angle BCA \approx 39.23^\circ$$ #### **Angle $$\angle CAB$$:** We can find the remaining angle $$\angle CAB$$ by using the fact that the sum of the angles in any triangle is $$180^\circ$$: $$\angle CAB = 180^\circ - \angle ABC - \angle BCA$$ $$\angle CAB = 180^\circ - 59.46^\circ - 39.23^\circ$$ $$\angle CAB \approx 81.31^\circ$$ ### Final Results: - **Side AB** = 9 units - **Side BC** = $$\sqrt{34} \approx 5.83$$ units - **Side AC** = $$\sqrt{61} \approx 7.81$$ units - **Angle $$\angle ABC$$ ≈ 59.46°** - **Angle $$\angle BCA$$ ≈ 39.23°** - **Angle $$\angle CAB$$ ≈ 81.31°** Let me know if you need further clarification or details! ### Related Questions: 1. How do you apply the distance formula to find the length of a side in a triangle? 2. What is the Law of Cosines, and when is it used in trigonometry? 3. How can you find the area of a triangle when the vertices are known? 4. What is the Law of Sines, and how does it relate to the Law of Cosines? 5. Can you solve for the angles of a triangle if only the lengths of the sides are known? **Tip:** Remember that the sum of the interior angles of any triangle is always $$180^\circ$$, so if you know two angles, you can easily find the third one.