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### Problem 1: Finding the Angle at Vertex \( B \)
The given triangle has vertices at:
- \( A(3, 5) \)
- \( B(-1, -5) \)
- \( C(4, 3) \)

To find the angle at vertex \( B \), we can use the dot product formula for vectors. Let's first calculate the vectors \( \overrightarrow{BA} \) and \( \overrightarrow{BC} \):
- \( \overrightarrow{BA} = A - B = (3 - (-1), 5 - (-5)) = (4, 10) \)
- \( \overrightarrow{BC} = C - B = (4 - (-1), 3 - (-5)) = (5, 8) \)

The dot product \( \overrightarrow{BA} \cdot \overrightarrow{BC} \) is given by:
\[ \overrightarrow{BA} \cdot \overrightarrow{BC} = (4)(5) + (10)(8) = 20 + 80 = 100 \]

The magnitudes of \( \overrightarrow{BA} \) and \( \overrightarrow{BC} \) are:
\[ |\overrightarrow{BA}| = \sqrt{4^2 + 10^2} = \sqrt{16 + 100} = \sqrt{116} = 2\sqrt{29} \]
\[ |\overrightarrow{BC}| = \sqrt{5^2 + 8^2} = \sqrt{25 + 64} = \sqrt{89} \]

Using the dot product formula:
\[ \overrightarrow{BA} \cdot \overrightarrow{BC} = |\overrightarrow{BA}| |\overrightarrow{BC}| \cos \theta \]
\[ 100 = (2\sqrt{29})(\sqrt{89}) \cos \theta \]
\[ \cos \theta = \frac{100}{2\sqrt{29} \cdot \sqrt{89}} = \frac{100}{2\sqrt{2581}} = \frac{50}{\sqrt{2581}} \]
\[ \theta = \cos^{-1} \left( \frac{50}{\sqrt{2581}} \right) \]

Thus, the angle at vertex \( B \) can be found using the inverse cosine function.

### Problem 2: Verifying the Pythagorean Theorem
To verify that the given triangle is a right triangle using the Pythagorean theorem, we need to check the lengths of the sides:
- \( AB = |\overrightarrow{BA}| = 2\sqrt{29} \)
- \( BC = |\overrightarrow{BC}| = \sqrt{89} \)
- \( AC \)

Calculate \( AC \):
\[ \overrightarrow{AC} = C - A = (4 - 3, 3 - 5) = (1, -2) \]
\[ AC = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

Now, verify the Pythagorean theorem:
\[ AB^2 + BC^2 = (2\sqrt{29})^2 + (\sqrt{89})^2 = 4 \cdot 29 + 89 = 116 + 89 = 205 \]
\[ AC^2 = (\sqrt{5})^2 = 5 \]

Since \( AB^2 + BC^2 \neq AC^2 \), the given points do not form a right triangle.

### Problem 3: Finding the Area of Triangle \( ABC \)
The area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) is given by:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

For vertices \( A(3, 5) \), \( B(-1, -5) \), \( C(4, 3) \):
\[ \text{Area} = \frac{1}{2} \left| 3(-5 - 3) + (-1)(3 - 5) + 4(5 + 5) \right| \]
\[ = \frac{1}{2} \left| 3(-8) + (-1)(-2) + 4(10) \right| \]
\[ = \frac{1}{2} \left| -24 + 2 + 40 \right| \]
\[ = \frac{1}{2} \left| 18 \right| \]
\[ = 9 \]

Therefore, the area of triangle \( ABC \) is 9 square units.

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### Related Questions:
1. How do you find the angle at a different vertex in a triangle using vectors?
2. What is the process for verifying if any given triangle is a right triangle?
3. How do you calculate the area of a triangle using determinant formulas?
4. What are some other methods to find the area of a triangle given coordinates?
5. How can the law of cosines be applied to find angles in a triangle?
6. What is the significance of the dot product in determining angles between vectors?
7. How do you derive the formula for the magnitude of a vector?
8. What are the properties of a right triangle in the context of coordinate geometry?

### Tip:
When working with vectors to find angles, always ensure you carefully compute the dot product and magnitudes, as precision is key to obtaining accurate results.

Explore how to find angles, verify triangle properties, and calculate area using vectors and coordinate geometry. Step-by-step solutions are provided for problems involving triangles with given vertices. Suitable for high school students studying coordinate geometry.

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