A man wandering in the desert walks 5.7 miles in the direction S 26 degrees﻿﻿ W. He then turns ﻿﻿90 degrees and walks 9 miles in the direction N ﻿﻿ W. At that time, how far is he from his starting point, and what is his bearing from his starting point?

We can solve this problem using vector analysis and trigonometry. The man walks in two legs, and we need to determine his distance from the starting point (which will form a triangle) and his final bearing.

### Step 1: Breakdown of the directions
- First, the man walks **5.7 miles** in the direction **S 26° W**. This means the angle is 26 degrees west of the south.
- Then, he turns **90°** and walks **9 miles** in the direction **N W**, which means 45 degrees west of north.

We want to determine:
1. The straight-line distance from his final position back to his starting point (i.e., the hypotenuse of a triangle).
2. His bearing from the starting point.

### Step 2: Decompose the walks into components
We'll break each leg of his walk into its north-south (vertical) and east-west (horizontal) components using trigonometry.

#### Leg 1: 5.7 miles at S 26° W
- **Southward component**: \( 5.7 \times \cos(26^\circ) \)
- **Westward component**: \( 5.7 \times \sin(26^\circ) \)

Calculating:
\[
\text{Southward component} = 5.7 \times \cos(26^\circ) \approx 5.7 \times 0.8988 = 5.122 \text{ miles south}
\]
\[
\text{Westward component} = 5.7 \times \sin(26^\circ) \approx 5.7 \times 0.4384 = 2.500 \text{ miles west}
\]

#### Leg 2: 9 miles at N W (45 degrees west of north)
- **Northward component**: \( 9 \times \cos(45^\circ) \)
- **Westward component**: \( 9 \times \sin(45^\circ) \)

Since \( \cos(45^\circ) = \sin(45^\circ) \approx 0.7071 \), we can calculate:

\[
\text{Northward component} = 9 \times 0.7071 = 6.364 \text{ miles north}
\]
\[
\text{Westward component} = 9 \times 0.7071 = 6.364 \text{ miles west}
\]

### Step 3: Total displacement in the north-south and east-west directions
Now we sum the north-south and east-west components.

- **North-south direction**:
  - He walked 5.122 miles south in the first leg and 6.364 miles north in the second leg.
  - Net displacement: \( 6.364 - 5.122 = 1.242 \text{ miles north} \)

- **East-west direction**:
  - He walked 2.500 miles west in the first leg and 6.364 miles west in the second leg.
  - Net displacement: \( 2.500 + 6.364 = 8.864 \text{ miles west} \)

### Step 4: Calculate the straight-line distance (resultant vector)
The man's straight-line distance from his starting point is the hypotenuse of the right triangle formed by the north-south and east-west displacements. Use the Pythagorean theorem:

\[
d = \sqrt{(1.242)^2 + (8.864)^2}
\]

Calculating:

\[
d = \sqrt{1.542 + 78.573} = \sqrt{80.115} \approx 8.951 \text{ miles}
\]

### Step 5: Determine the bearing
The bearing

A man wandering in the desert walks 5.7 miles in the direction S 26° W. He then turns 90° and walks 9 miles in the direction NW. At that time, how far is he from his starting point, and what is his bearing from his starting point?

Solve how far a man is from his starting point after walking 5.7 miles S 26° W and 9 miles NW. This trigonometry problem uses vector components, the Pythagorean theorem, and bearing calculation, suitable for students in Grades 9-12.

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