3 Vector Worksheet PDF | PDF | Trigonometric Functions | Motion ... - Free Printable

Problem Overview:

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

We will solve each part step by step.

---

Section 1: Using the Pythagorean Theorem to Solve for \( x \)

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:
\[
8^2 + x^2 = 10^2
\]
\[
64 + x^2 = 100
\]
\[
x^2 = 36
\]
\[
x = \sqrt{36} = 6
\]

#### 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 the Sine, Cosine, and Tangent of the Angle \( \theta \)

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. \( \sin \theta = \frac{8}{17}, \cos \theta = \frac{15}{17}, \tan \theta = \frac{8}{15} \)

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