The answer key says it's 2.6 or 10, there's two triangles 

Given the two possible answers, it indicates that there are two possible solutions for the distance between Martina and Carl due to the ambiguous case in trigonometry known as the "SSA" (Side-Side-Angle) condition, where two sides and a non-included angle are known. 

Let's solve for both possible distances using the Law of Cosines and the Law of Sines.

1. **Using the Law of Cosines:**

   As derived earlier, one possible solution for \( c \) is:

   \[
   c^2 = 7.8^2 + 5.9^2 - 2 \times 7.8 \times 5.9 \times \cos(36^\circ)
   \]
   \[
   c^2 = 95.65 - 74.236
   \]
   \[
   c^2 = 21.414
   \]
   \[
   c = \sqrt{21.414} \approx 4.6 \, \text{m}
   \]

   It appears that there was a calculation mistake here, because neither 2.6 nor 10 were obtained. This suggests that we need to examine the possible ambiguous cases more closely with the Law of Sines.

2. **Using the Law of Sines:**

   The Law of Sines states:

   \[
   \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}
   \]

   First, we calculate the possible angle \( B \):

   \[
   \frac{7.8}{\sin(B)} = \frac{5.9}{\sin(36^\circ)}
   \]
   \[
   \sin(B) = \frac{7.8 \times \sin(36^\circ)}{5.9}
   \]

   Calculate \( \sin(36^\circ) \):
   \[
   \sin(36^\circ) \approx 0.588
   \]

   Therefore:
   \[
   \sin(B) = \frac{7.8 \times 0.588}{5.9} \approx 0.777
   \]

   Since \( \sin(B) \approx 0.777 \), the angle \( B \) can be:
   \[
   B = \sin^{-1}(0.777) \approx 51^\circ \text{ or } 129^\circ
   \]

   Now, let's consider both cases for angle \( B \):

   - **Case 1: \( B = 51^\circ \)**

     Then angle \( A \):
     \[
     A = 180^\circ - 36^\circ - 51^\circ = 93^\circ
     \]

     Using the Law of Sines again to find \( c \):
     \[
     \frac{c}{\sin(93^\circ)} = \frac{5.9}{\sin(36^\circ)}
     \]
     \[
     c = \frac{5.9 \times \sin(93^\circ)}{0.588} \approx \frac{5.9 \times 0.998}{0.588} \approx 10 \, \text{m}
     \]

   - **Case 2: \( B = 129^\circ \)**

     Then angle \( A \):
     \[
     A = 180^\circ - 36^\circ - 129^\circ = 15^\circ
     \]

     Using the Law of Sines again to find \( c \):
     \[
     \frac{c}{\sin(15^\circ)} = \frac{5.9}{\sin(36^\circ)}
     \]
     \[
     c = \frac{5.9 \times \sin(15^\circ)}{0.588} \approx \frac{5.9 \times 0.259}{0.588} \approx 2.6 \, \text{m}
     \]

Therefore, the two possible distances between Martina and Carl are approximately \( 2.6 \) meters and \( 10 \) meters.

Do you want details or have any questions?

Here are 5 questions you might want to ask:
1. How does the ambiguous case of the Law of Sines affect solving triangles?
2. Can you explain the derivation and application of the Law of Cosines?
3. What are common mistakes to avoid when solving SSA triangle problems?
4. How do you determine which trigonometric function to use in different problems?
5. Can you provide more examples of using the Law of Sines and Law of Cosines?

**Tip:** When solving SSA triangle problems, always check for the ambiguous case to ensure all possible solutions are considered.

Learn how to solve a trigonometry problem involving the ambiguous case of the Law of Sines. Understand how to apply the Law of Cosines and Law of Sines to determine multiple solutions for triangle side lengths. Suitable for high school students studying trigonometry.

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