Your friend has slipped and fallen. To help her up, you pull with a force F, as the drawing shows. The vertical component of this force is 130 newtons, and the horizontal component is 150 newtons. Find (a) the magnitude of F and (b) the angle 0.

To solve this problem, we'll use the principles of vector addition and trigonometry.

### Given:
- Vertical component of the force (\(F_y\)) = 130 N
- Horizontal component of the force (\(F_x\)) = 150 N

### Part (a): Find the magnitude of the force \( F \)

The magnitude of the force \( F \) can be found using the Pythagorean theorem because the force components form a right triangle.

\[
F = \sqrt{F_x^2 + F_y^2}
\]

Substitute the values:

\[
F = \sqrt{(150)^2 + (130)^2}
\]

\[
F = \sqrt{22500 + 16900} = \sqrt{39400} \approx 198.49 \, \text{N}
\]

### Part (b): Find the angle \( \theta \)

The angle \( \theta \) that the force makes with the horizontal can be found using the tangent function, which relates the angle to the opposite and adjacent sides of the triangle.

\[
\tan(\theta) = \frac{F_y}{F_x}
\]

\[
\theta = \tan^{-1}\left(\frac{130}{150}\right)
\]

\[
\theta = \tan^{-1}(0.867) \approx 40.93^\circ
\]

### Summary:
- **(a)** The magnitude of the force \( F \) is approximately **198.49 N**.
- **(b)** The angle \( \theta

Learn how to calculate the magnitude and angle of a force using vector addition and trigonometry. Solve the problem involving force components of 130 N vertically and 150 N horizontally. Suitable for high school students studying vectors and trigonometry.

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