Question

use Pythagorean theorem to find the side angle
a is -7,6
b is -2,-6
c is 10,-1

Answer 1

To find the angle at vertex \( A \) (which is point \( A = (-7, 6) \)) using the Pythagorean theorem, we first need to find the lengths of the sides of the triangle formed by points \( A \), \( B \), and \( C \).

Let:
- Point \( A = (-7, 6) \)
- Point \( B = (-2, -6) \)
- Point \( C = (10, -1) \)

### Step 1: Calculate the lengths of the sides.

1. Length \( AB \):
\[
AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{((-2) - (-7))^2 + ((-6) - 6)^2}
\]
\[
= \sqrt{(-2 + 7)^2 + (-6 - 6)^2} = \sqrt{(5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13
\]

2. Length \( AC \):
\[
AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} = \sqrt{(10 - (-7))^2 + ((-1) - 6)^2}
\]
\[
= \sqrt{(10 + 7)^2 + (-1 - 6)^2} = \sqrt{(17)^2 + (-7)^2} = \sqrt{289 + 49} = \sqrt{338}
\]

3. Length \( BC \):
\[
BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} = \sqrt{(10 - (-2))^2 + ((-1) - (-6))^2}
\]
\[
= \sqrt{(10 + 2)^2 + (-1 + 6)^2} = \sqrt{(12)^2 + (5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13
\]

### Step 2: Assign the lengths to sides

- \( a = BC = 13 \)
- \( b = AC = \sqrt{338} \)
- \( c = AB = 13 \)

### Step 3: Use the cosine rule to find the angle \( A \)

Using the cosine rule:
\[
\cos A = \frac{b^2 + c^2 - a^2}{2bc}
\]

Plugging in the values:
\[
b = \sqrt{338}, \quad c = 13, \quad a = 13
\]
\[
\cos A = \frac{(\sqrt{338})^2 + 13^2 - 13^2}{2 \cdot \sqrt{338} \cdot 13} = \frac{338 + 169 - 169}{2 \cdot \sqrt{338} \cdot 13} = \frac{338}{2 \cdot \sqrt{338} \cdot 13}
\]
\[
= \frac{169}{\sqrt{338} \cdot 13} = \frac{169}{13 \sqrt{338}} = \frac{13}{\sqrt{338}}
\]

### Step 4: Find angle \( A \)

To find \( A \), take the arccosine:
\[
A = \cos^{-1} \left( \frac{13}{\sqrt{338}} \right)
\]

### Step 5: Calculate

Using a calculator:
- Calculate \( \sqrt{338} \approx 18.38 \).
- Then \( \frac{13}{\sqrt{338}} \approx \frac{13}{18.38} \approx 0.707 \).

Finally, find the angle:
\[
A \approx \cos^{-1}(0.707) \approx 45^\circ.
\]

So, the angle at point \( A \) is approximately \( 45^\circ \).