An athlete participating in a paragliding competition was to reach the landing area B located 34 miles east of the starting position A. Due to the north wind, the athlete deviated from the intended course and landed directly south of B. The bearing from the starting position to the actual landing point is S 79.9° E. How far is the landing point from B? Round your answer to two decimal places.





We can solve this problem using trigonometry. Specifically, this is a scenario where we can apply the law of sines.

### Step-by-step Solution:

1. **Identify the triangle:**
   - The starting position \( A \) is the vertex.
   - The intended landing position \( B \) is 34 miles east of \( A \).
   - The actual landing point \( C \) is directly south of \( B \).
   - The bearing from \( A \) to \( C \) is \( S 79.9^\circ E \). This means the angle between the eastward direction from \( A \) to \( B \) and the direction from \( A \) to \( C \) is \( 79.9^\circ \).

2. **Set up the triangle:**
   - You have a triangle where:
     - Side \( AB = 34 \) miles.
     - Angle \( \angle BAC = 79.9^\circ \).
     - Angle \( \angle ABC = 90^\circ \) (since the athlete lands directly south of \( B \)).

3. **Find the third angle:**
   Since the sum of angles in a triangle is \( 180^\circ \), the third angle \( \angle ACB \) can be found as:
   \[
   \angle ACB = 180^\circ - 90^\circ - 79.9^\circ = 10.1^\circ
   \]

4. **Use the law of sines:**
   The law of sines states:
   \[
   \frac{AB}{\sin(\angle ACB)} = \frac{AC}{\sin(\angle ABC)}
   \]
   Substituting the known values:
   \[
   \frac{34}{\sin(10.1^\circ)} = \frac{AC}{\sin(90^\circ)}
   \]
   Since \( \sin(90^\circ) = 1 \), this simplifies to:
   \[
   AC = \frac{34}{\sin(10.1^\circ)}
   \]
   Now, calculate \( \sin(10.1^\circ) \) and solve for \( AC \).

Let's compute this.The distance from the actual landing point to point \( B \) is approximately \( 193.88 \) miles.

Would you like further clarification or a breakdown of any specific step?

Here are some related questions:
1. What is the law of sines and how does it work in solving triangles?
2. How does the bearing system work, especially in navigation problems?
3. Can we use other trigonometric rules, such as the law of cosines, in this problem?
4. What are some practical applications of trigonometry in navigation and sports like paragliding?
5. How do angles in bearings relate to cardinal directions (north, east, south, west)?

**Tip**: When working with bearings, always measure angles clockwise from the north (unless otherwise stated). This helps avoid confusion with directions.

An athlete participating in a paragliding competition was to reach the landing area B located 34 miles east of the starting position A. Due to the north wind, the athlete deviated from the intended course and landed directly south of B. The bearing from the starting position to the actual landing point is S 79.9° E. How far is the landing point from B? Round your answer to two decimal places.

This problem involves applying trigonometry to find the distance between the actual and intended landing points of a paraglider. The solution uses the law of sines, involving a 34-mile distance, bearings, and angles, resulting in a final distance of 193.88 miles. Suitable for students in Grades 10-12.

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