You are heading to Beech Mountain for a ski trip. Unfortunately, state road 105 in North Carolina is blocked off due to a chemical spill. You have to get to Tynecastle Highway which leads to the resort at which you are staying. NC-105 would get you to Tynecastle Hwy in 12.8 miles. The detour begins with a 18 degree veer off onto a road that runs through the local city. After 6 miles, there is another turn that leads to Tynecastle Hwy. Assuming that both roads on the detour are straight, how many extra miles are you traveling to reach your destination? (Hint: x < 12.8)


To solve this problem, we can break it down into the following steps:

1. **Understand the Detour Path**:
   - The original path along NC-105 was 12.8 miles.
   - The detour begins with a veer of 18 degrees off from NC-105.
   - After 6 miles on the detour, there is another turn leading directly to Tynecastle Hwy.

2. **Forming a Triangle**:
   - The original straight path along NC-105 is the base of the triangle, 12.8 miles.
   - The first leg of the detour is 6 miles, which is one side of the triangle.
   - We need to calculate the length of the second leg of the detour, which will take you from the end of the 6-mile road back to Tynecastle Hwy.

3. **Applying the Law of Cosines**:
   The law of cosines is useful here since we know two sides of the triangle and the angle between them:

   \[
   c^2 = a^2 + b^2 - 2ab \cos(C)
   \]

   Where:
   - \( a = 6 \) miles (first detour leg)
   - \( b = 12.8 \) miles (straight route along NC-105)
   - \( C = 18^\circ \) (the angle between the original path and the first detour leg)
   - \( c \) is the distance from the end of the first leg of the detour to Tynecastle Hwy (the second detour leg).

   Plugging in the values:

   \[
   c^2 = 6^2 + 12.8^2 - 2(6)(12.8)\cos(18^\circ)
   \]

   Let's calculate \( c \):

   \[
   c^2 = 36 + 163.84 - 153.6 \cdot \cos(18^\circ)
   \]

   \[
   c^2 = 199.84 - 153.6 \cdot \cos(18^\circ)
   \]

   Calculating \( \cos(18^\circ) \approx 0.95106 \):

   \[
   c^2 = 199.84 - 153.6 \times 0.95106
   \]

   \[
   c^2 = 199.84 - 146.09
   \]

   \[
   c^2 = 53.75
   \]

   \[
   c \approx \sqrt{53.75} \approx 7.33 \text{ miles}
   \]

4. **Calculate the Extra Distance**:
   The total distance traveled on the detour is \( 6 + 7.33 = 13.33 \) miles.

   The extra distance traveled is:

   \[
   13.33 - 12.8 = 0.53 \text{ miles}
   \]

So, you are traveling **0.53 extra miles** to reach your destination.

Would you like any further details or have any questions?

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Here are 5 related questions to expand your understanding:

1. How would the extra distance change if the angle of veer was 25 degrees instead of 18 degrees?
2. How does the Law of Cosines compare to the Pythagorean Theorem when solving for distances in triangles?
3. What if the first detour leg was 10 miles long? How would that affect the extra distance?
4. Can the Law of Sines be used instead of the Law of Cosines in this scenario? Why or why not?
5. How would the solution change if the roads were not straight?

**Tip:** When solving problems involving angles and distances, drawing a diagram can help visualize the relationships and make applying trigonometric laws easier.

Learn how to calculate the extra miles traveled on a detour due to a chemical spill blocking NC-105 in North Carolina. This problem involves trigonometric principles, including the Law of Cosines, to determine the additional distance traveled. Suitable for advanced high school students.

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