37. Go The components of vector A are A, and A, (both positive), and the angle that it makes with respect to the positive x axis is 0. Find the angle 0 if the components of the displacement vector A are (a) A, = 12 m and A, 12 m, (b) A, = 17 m and Ay 12 m, and (c) A, A, = 17 m. = 12 m and =

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**Question 37: Vector Components and Angles**

**Problem Statement:**
The components of vector **A** are \( A_x \) and \( A_y \) (both positive), and the angle that it makes with respect to the positive x axis is \( \theta \). Find the angle \( \theta \) if the components of the displacement vector **A** are:

**(a)** \( A_x = 12 \, \text{m} \) and \( A_y = 12 \, \text{m} \)

**(b)** \( A_x = 17 \, \text{m} \) and \( A_y = 12 \, \text{m} \)

**(c)** \( A_x = 12 \, \text{m} \) and \( A_y = 17 \, \text{m} \)

**Explanation:**
### Solution Overview:

To find the angle \( \theta \) made with the positive x-axis, you can use the trigonometric relation involving the arctangent function:

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

### Detailed Solutions:

1. **Case (a):**
   Given \( A_x = 12 \, \text{m} \) and \( A_y = 12 \, \text{m} \),

   \[
   \theta = \tan^{-1} \left( \frac{12}{12} \right) = \tan^{-1}(1) = 45^\circ
   \]

2. **Case (b):**
   Given \( A_x = 17 \, \text{m} \) and \( A_y = 12 \, \text{m} \),

   \[
   \theta = \tan^{-1} \left( \frac{12}{17} \right)
   \]

   Using a calculator or trigonometric tables to find the inverse tangent,

   \[
   \theta \approx 35.5^\circ
   \]

3. **Case (c):**
   Given \( A_x = 12 \, \text{m} \) and \( A_y = 17 \, \text{m} \),

   \[
   \theta = \tan^{-1} \left( \frac{17}{12}
Transcribed Image Text:**Question 37: Vector Components and Angles** **Problem Statement:** The components of vector **A** are \( A_x \) and \( A_y \) (both positive), and the angle that it makes with respect to the positive x axis is \( \theta \). Find the angle \( \theta \) if the components of the displacement vector **A** are: **(a)** \( A_x = 12 \, \text{m} \) and \( A_y = 12 \, \text{m} \) **(b)** \( A_x = 17 \, \text{m} \) and \( A_y = 12 \, \text{m} \) **(c)** \( A_x = 12 \, \text{m} \) and \( A_y = 17 \, \text{m} \) **Explanation:** ### Solution Overview: To find the angle \( \theta \) made with the positive x-axis, you can use the trigonometric relation involving the arctangent function: \[ \theta = \tan^{-1} \left( \frac{A_y}{A_x} \right) \] ### Detailed Solutions: 1. **Case (a):** Given \( A_x = 12 \, \text{m} \) and \( A_y = 12 \, \text{m} \), \[ \theta = \tan^{-1} \left( \frac{12}{12} \right) = \tan^{-1}(1) = 45^\circ \] 2. **Case (b):** Given \( A_x = 17 \, \text{m} \) and \( A_y = 12 \, \text{m} \), \[ \theta = \tan^{-1} \left( \frac{12}{17} \right) \] Using a calculator or trigonometric tables to find the inverse tangent, \[ \theta \approx 35.5^\circ \] 3. **Case (c):** Given \( A_x = 12 \, \text{m} \) and \( A_y = 17 \, \text{m} \), \[ \theta = \tan^{-1} \left( \frac{17}{12}
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