Final answer:
To find the angles of the triangle with vertices A (0, -2), B (6, 4), and C (3, 4), we can use the distance formula to find the lengths of the sides of the triangle. Once we have the lengths of the sides, we can use the Law of Cosines to find the angles.
Step-by-step explanation:
To find the angles of the triangle with vertices A (0, -2), B (6, 4), and C (3, 4), we can use the distance formula to find the lengths of the sides of the triangle. Once we have the lengths of the sides, we can use the Law of Cosines to find the angles.
Let's label the sides of the triangle as a, b, and c, with opposite angles A, B, and C, respectively. The distance between points A and B is the side c, the distance between points B and C is the side a, and the distance between points C and A is the side b.
Using the distance formula, we find that c = √[(6-0)² + (4-(-2))²] = √(36 + 36) = √72 = 6√2. Similarly, we find that a = √[(3-6)² + (4-4)²] = √9 = 3, and b = √[(3-0)² + (4-(-2))²] = √49 = 7.
Now we can use the Law of Cosines to find the angles:
cos(A) = (b² + c² - a²) / (2bc)
cos(B) = (a² + c² - b²) / (2ac)
cos(C) = (a² + b² - c²) / (2ab)
Plugging in the values, we can calculate the angles:
cos(A) = (3² + (6√2)² - 7²) / (2 * 3 * (6√2)) = 67/18√2
cos(B) = (7² + (6√2)² - 3²) / (2 * 7 * (6√2)) = 2/3√2
cos(C) = (3² + 7² - (6√2)²) / (2 * 3 * 7) = 43/9√2
Finally, we can use the inverse cosine function to find the angles:
A ≈ acos(67/18√2)
B ≈ acos(2/3√2)
C ≈ acos(43/9√2)