**Vector Decomposition and Trigonometry in Navigation****Problem Statement:**A person walks \(26.0^\circ\) north of east for \(3.70\) km. How far due north and how far due east would she have to walk to arrive at the same location?**Interactive Input Fields:**- **North:** \( \_\_\_\_\_\_\_\_ \) km- **East:** \( \_\_\_\_\_\_\_\_ \) km**Guidance Section:**- **Need Help?** Click the "Read It" button.**Detailed Explanation:**This exercise involves breaking down the movement into its vector components using trigonometry. The person’s walk is along a line at an angle relative to the east direction. To find out how much of this walk contributes to the northern and eastern directions individually, we use the sine and cosine functions.Given:- The angle \( \theta = 26.0^\circ \) north of east- The total distance \( d = 3.70 \) kilometersWe can find the northern and eastern components of the walk by using the following trigonometric relationships:- **North Component (opposite side, using sine):** \[ \text{North} = d \cdot \sin(\theta) \]- **East Component (adjacent side, using cosine):** \[ \text{East} = d \cdot \cos(\theta) \]This decomposition allows us to understand the exact northward and eastward distances traveled, which can be useful in navigation and mapping exercises.**Example Calculation:**1. Calculate the northward distance: \[ \text{North} = 3.70 \times \sin(26.0^\circ) \]2. Calculate the eastward distance: \[ \text{East} = 3.70 \times \cos(26.0^\circ) \]By inputting the values into the fields provided, students can practice and verify their trigonometric calculations.**Note on Interactive Elements:**For additional help on solving this problem, click the "Read It" button in the "Need Help?" section for detailed guidance or examples.

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A person walks 26.0° north of east for 3.70 km. How far due north and how far due east would she have to walk to arrive at the same location? north km east km

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

**Vector Decomposition and Trigonometry in Navigation**

**Problem Statement:**
A person walks \(26.0^\circ\) north of east for \(3.70\) km. How far due north and how far due east would she have to walk to arrive at the same location?

**Interactive Input Fields:**
- **North:** \( \_\_\_\_\_\_\_\_ \) km
- **East:** \( \_\_\_\_\_\_\_\_ \) km

**Guidance Section:**
- **Need Help?** Click the "Read It" button.

**Detailed Explanation:**
This exercise involves breaking down the movement into its vector components using trigonometry. The person’s walk is along a line at an angle relative to the east direction. To find out how much of this walk contributes to the northern and eastern directions individually, we use the sine and cosine functions.

Given:
- The angle \( \theta = 26.0^\circ \) north of east
- The total distance \( d = 3.70 \) kilometers

We can find the northern and eastern components of the walk by using the following trigonometric relationships:
- **North Component (opposite side, using sine):**
\[ \text{North} = d \cdot \sin(\theta) \]
- **East Component (adjacent side, using cosine):**
\[ \text{East} = d \cdot \cos(\theta) \]

This decomposition allows us to understand the exact northward and eastward distances traveled, which can be useful in navigation and mapping exercises.

**Example Calculation:**
1. Calculate the northward distance:
\[ \text{North} = 3.70 \times \sin(26.0^\circ) \]
2. Calculate the eastward distance:
\[ \text{East} = 3.70 \times \cos(26.0^\circ) \]

By inputting the values into the fields provided, students can practice and verify their trigonometric calculations.

**Note on Interactive Elements:**
For additional help on solving this problem, click the "Read It" button in the "Need Help?" section for detailed guidance or examples.

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