We are tasked with finding the angles of triangle ABC, given the vertices $A(-7, 4)$, $B(-3, -2)$, and $C(-1, 0)$. Let's go step by step to determine the angle measures and deduce information about the triangle.

Steps to Solve:

1. Find the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ This will give us the lengths of sides $AB$, $BC$, and $AC$.

2. Use the Law of Cosines to find the angles: $\cos(\theta) = \frac{b^2 + c^2 - a^2}{2bc}$ where $a$, $b$, and $c$ are the side lengths.

3. Sum of the angles in a triangle is 180°, so we can use this property to check our results.

Let me first calculate the side lengths and then proceed with determining the angles.The side lengths of triangle ABC are:

• $AB = 7.21$ units
• $BC = 2.83$ units
• $AC = 7.21$ units

The measures of the angles are:

• $\angle A = 22.62^\circ$
• $\angle B = 78.69^\circ$
• $\angle C = 78.69^\circ$

Deduction:

Since two angles of the triangle (B and C) are equal, triangle ABC is isosceles.

Would you like further details or explanations?

Here are 5 related questions to consider:

1. How would the process change if one vertex was in 3D space?
2. What is the significance of the Law of Cosines in non-right triangles?
3. Can you use trigonometry to solve a triangle without knowing side lengths?
4. How can you verify that a triangle is isosceles given only its vertices?
5. What properties differentiate an isosceles triangle from a scalene triangle?

Tip: When working with isosceles triangles, knowing two equal sides or angles simplifies solving for other properties!