Let's solve the given problems using the specified position vectors.

Given:

The position vectors are:

• $\vec{a} = 3\hat{i} + 5\hat{j}$
• $\vec{b} = \hat{i} - 5\hat{j}$
• $\vec{c} = 4\hat{i} + 3\hat{j}$

Problem 1: Find $\angle ABC$

To find the angle at vertex $B$, we use the dot product formula. Let's calculate the vectors $\vec{BA}$ and $\vec{BC}$:

• $\vec{BA} = \vec{a} - \vec{b} = (3\hat{i} + 5\hat{j}) - (\hat{i} - 5\hat{j}) = (3 - 1)\hat{i} + (5 - (-5))\hat{j} = 2\hat{i} + 10\hat{j}$
• $\vec{BC} = \vec{c} - \vec{b} = (4\hat{i} + 3\hat{j}) - (\hat{i} - 5\hat{j}) = (4 - 1)\hat{i} + (3 - (-5))\hat{j} = 3\hat{i} + 8\hat{j}$

The dot product $\vec{BA} \cdot \vec{BC}$ is: $\vec{BA} \cdot \vec{BC} = (2\hat{i} + 10\hat{j}) \cdot (3\hat{i} + 8\hat{j}) = (2 \cdot 3) + (10 \cdot 8) = 6 + 80 = 86$

The magnitudes of $\vec{BA}$ and $\vec{BC}$ are: $|\vec{BA}| = \sqrt{2^2 + 10^2} = \sqrt{4 + 100} = \sqrt{104} = 2\sqrt{26}$ $|\vec{BC}| = \sqrt{3^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73}$

Using the dot product formula: $\cos \theta = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}| |\vec{BC}|} = \frac{86}{2\sqrt{26} \cdot \sqrt{73}} = \frac{86}{2\sqrt{1898}} = \frac{43}{\sqrt{1898}}$

Thus, the angle $\theta$ is: $\theta = \cos^{-1} \left( \frac{43}{\sqrt{1898}} \right)$

Problem 2: Verify Pythagorean Theorem

To verify that the triangle is a right triangle, we calculate the squares of the lengths of all sides:

• $|\vec{AB}| = |\vec{BA}| = 2\sqrt{26}$
• ( |\vec{BC}| = \sqrt{