Let's analyze the provided questions step-by-step.

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.