Vector Worksheet with Pythagorean Theorem and Trigonometry Problems

Problem Overview:

The worksheet involves two main sections:
1. Using the Pythagorean Theorem to solve for \( x \) in right triangles.
2. Finding the sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)) of an angle \( \theta \) in right triangles.

We will solve each part step by step.

---

Section 1: Using the Pythagorean Theorem

The Pythagorean Theorem states:
\[
a^2 + b^2 = c^2
\]
where \( a \) and \( b \) are the legs of the right triangle, and \( c \) is the hypotenuse.

#### Problem 1:
- Given: \( a = 5 \), \( b = 5 \), \( c = x \)
- Apply the Pythagorean Theorem:
\[
5^2 + 5^2 = x^2
\]
\[
25 + 25 = x^2
\]
\[
50 = x^2
\]
\[
x = \sqrt{50} = 5\sqrt{2}
\]

#### Problem 2:
- Given: \( a = 8 \), \( b = 12 \), \( c = x \)
- Apply the Pythagorean Theorem:
\[
8^2 + 12^2 = x^2
\]
\[
64 + 144 = x^2
\]
\[
208 = x^2
\]
\[
x = \sqrt{208} = 4\sqrt{13}
\]

#### Problem 3:
- Given: \( a = 8 \), \( b = x \), \( c = 10 \)
- Apply the Pythagorean Theorem:
\[
8^2 + x^2 = 10^2
\]
\[
64 + x^2 = 100
\]
\[
x^2 = 100 - 64
\]
\[
x^2 = 36
\]
\[
x = \sqrt{36} = 6
\]

#### Problem 4:
- Given: \( a = 2 \), \( b = 4 \), \( c = x \)
- Apply the Pythagorean Theorem:
\[
2^2 + 4^2 = x^2
\]
\[
4 + 16 = x^2
\]
\[
20 = x^2
\]
\[
x = \sqrt{20} = 2\sqrt{5}
\]

---

Section 2: Finding \( \sin \theta \), \( \cos \theta \), and \( \tan \theta \)

For a right triangle:
- \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \)
- \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \)
- \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)

#### Problem 5:
- Given: \( \text{opposite} = 4 \), \( \text{adjacent} = 3 \), \( \text{hypotenuse} = 5 \)
- Calculate:
\[
\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}
\]
\[
\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3}{5}
\]
\[
\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{3}
\]

#### Problem 6:
- Given: \( \text{opposite} = 8 \), \( \text{adjacent} = 7 \), \( \text{hypotenuse} = \sqrt{8^2 + 7^2} = \sqrt{64 + 49} = \sqrt{113} \)
- Calculate:
\[
\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{8}{\sqrt{113}}
\]
\[
\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{7}{\sqrt{113}}
\]
\[
\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{8}{7}
\]

#### Problem 7:
- Given: \( \text{opposite} = 5 \), \( \text{adjacent} = 12 \), \( \text{hypotenuse} = 13 \)
- Calculate:
\[
\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}
\]
\[
\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}
\]
\[
\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{12}
\]

#### Problem 8:
- Given: \( \text{opposite} = 8 \), \( \text{adjacent} = 15 \), \( \text{hypotenuse} = 17 \)
- Calculate:
\[
\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{8}{17}
\]
\[
\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{15}{17}
\]
\[
\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{8}{15}
\]

---

Final Answers:

1. \( x = 5\sqrt{2} \)
2. \( x = 4\sqrt{13} \)
3. \( x = 6 \)
4. \( x = 2\sqrt{5} \)

5. \( \sin \theta = \frac{4}{5}, \cos \theta = \frac{3}{5}, \tan \theta = \frac{4}{3} \)
6. \( \sin \theta = \frac{8}{\sqrt{113}}, \cos \theta = \frac{7}{\sqrt{113}}, \tan \theta = \frac{8}{7} \)
7. \( \sin \theta = \frac{5}{13}, \cos \theta = \frac{12}{13}, \tan \theta = \frac{5}{12} \)
8. \( \sin \theta = \frac{8}{17}, \cos \theta = \frac{15}{17}, \tan \theta = \frac{8}{15} \)

\[
\boxed{
\begin{array}{l}
1. 5\sqrt{2} \\
2. 4\sqrt{13} \\
3. 6 \\
4. 2\sqrt{5} \\
5. \sin \theta = \frac{4}{5}, \cos \theta = \frac{3}{5}, \tan \theta = \frac{4}{3} \\
6. \sin \theta = \frac{8}{\sqrt{113}}, \cos \theta = \frac{7}{\sqrt{113}}, \tan \theta = \frac{8}{7} \\
7. \sin \theta = \frac{5}{13}, \cos \theta = \frac{12}{13}, \tan \theta = \frac{5}{12} \\
8. \sin \theta = \frac{8}{17}, \cos \theta = \frac{15}{17}, \tan \theta = \frac{8}{15}
\end{array}
}
\]