To solve this problem, we will break it down into vectors and use trigonometry to find the distance Stella skied on the second leg of her trip.

1. Represent the first leg of the trip: Stella skied 8.0 km in the direction $\text{N}40^\circ \text{E}$. We can represent this in vector form. The north component (y-axis) and east component (x-axis) can be calculated using trigonometry. $\text{N}40^\circ \text{E} \text{ means } 40^\circ \text{ east of north}.$

   $x_1 = 8.0 \cos(40^\circ)$ $y_1 = 8.0 \sin(40^\circ)$

2. Calculate these components: $x_1 = 8.0 \cos(40^\circ) \approx 8.0 \times 0.7660 = 6.128 \text{ km}$ $y_1 = 8.0 \sin(40^\circ) \approx 8.0 \times 0.6428 = 5.1424 \text{ km}$

3. Represent the second leg of the trip: Stella then skied in the direction $\text{S}30^\circ \text{E}$. We can calculate this vector's components. $x_2 = d \cos(30^\circ)$ $y_2 = -d \sin(30^\circ)$

   where $d$ is the distance she skied on the second leg of her trip.

4. Combine the vectors: The total distance from her starting point to the cabin is 9.5 km. Combining the vectors from both legs: $x_1 + x_2 = 9.5 \cos(\theta)$ $y_1 + y_2 = 9.5 \sin(\theta)$

   Since $\theta = \arctan\left(\frac{y_1 + y_2}{x_1 + x_2}\right)$, and given that $9.5$ km is as the crow flies, we use the Pythagorean theorem to find the resultant vector: $\sqrt{(x_1 + x_2)^2 + (y_1 + y_2)^2} = 9.5 \text{ km}$

5. Solve for the distance $d$: $(6.128 + d \cos(30^\circ))^2 + (5.1424 - d \sin(30^\circ))^2 = 9.5^2$ Simplify the trigonometric functions: $\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$ $\sin(30^\circ) = \frac{1}{2} = 0.5$ So: $(6.128 + 0.866d)^2 + (5.1424 - 0.5d)^2 = 9.5^2$ $(6.128 + 0.866d)^2 + (5.1424 - 0.5d)^2 = 90.25$

   Expanding and simplifying this equation: $(6.128 + 0.866d)^2 = 37.905 + 10.604d + 0.749d^2$ $(5.1424 - 0.5d)^2 = 26.4384 - 5.1424d + 0.25d^2$ $37.905 + 10.604d + 0.749d^2 + 26.4384 - 5.1424d + 0.25d^2 = 90.25$ $0.999d^2 + 5.4616d + 64.3434 = 90.25$ $0.999d^2 + 5.4616d - 25.9066 = 0$

   Solving this quadratic equation for $d$: $d = \frac{-5.4616 \pm \sqrt{(5.4616)^2 - 4 \cdot 0.999 \cdot (-25.9066)}}{2 \cdot 0.999}$ $d = \frac{-5.4616 \pm \sqrt{29.8213 + 103.6204}}{1.998}$ $d = \frac{-5.4616 \pm \sqrt{133.4417}}{1.998}$ $d = \frac{-5.4616 \pm 11.553}{1.998}$

   Solving for the positive root: $d = \frac{6.0914}{1.998} \approx 3.05 \text{ km}$

So, the distance Stella traveled on the second leg of her trip is approximately $3.0$ km.

Would you like more details or have any questions?

Here are 8 related questions for further exploration:

1. How can vector addition be applied to solve real-world navigation problems?
2. What are the trigonometric identities and how are they used in navigation?
3. How does the Pythagorean theorem apply to vectors in two dimensions?
4. What is the significance of bearing angles in navigation?
5. How can the law of cosines be used in more complex navigation problems?
6. What are some real-life applications of vector addition besides navigation?
7. How do you convert between different forms of vector representation?
8. What are the methods for solving quadratic equations and their applications?

Tip: Always break down complex problems into smaller, manageable parts and solve step by step. This approach can simplify the problem and help avoid errors.