To solve this problem, we need to use the sine rule, which is expressed as:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Where:

• $a$, $b$, and $c$ are the lengths of the sides of the triangle.
• $A$, $B$, and $C$ are the angles opposite these sides.

In this triangle:

• We are given the side lengths and angles:
  • $a = 10 \, \text{cm}$, opposite angle $A = 59^\circ$,
  • $b = 13 \, \text{cm}$, opposite angle $B = 41^\circ$,
  • $c = 15 \, \text{cm}$, opposite the unknown angle $C$.

The equation asks us to complete the following expression: $\frac{10}{\sin(41^\circ)} = \frac{x}{\sin(59^\circ)}$

We need to solve for $x$. Using the sine rule: $x = \frac{10 \cdot \sin(59^\circ)}{\sin(41^\circ)}$

Now, let's calculate the value of $x$:

1. $\sin(59^\circ) \approx 0.8572$
2. $\sin(41^\circ) \approx 0.6561$

Now substitute these values:

$x = \frac{10 \cdot 0.8572}{0.6561} \approx \frac{8.572}{0.6561} \approx 13.06$

Thus, the number that goes in the box is approximately 13.06.

Would you like further clarification or more details?

Here are 5 related questions:

1. How does the sine rule apply to non-right triangles?
2. Can you use the cosine rule for this problem, and how does it differ?
3. What are the conditions for using the sine rule?
4. How would you calculate the third angle of this triangle?
5. What if the side lengths were different—how would that affect the equation?

Tip: When using trigonometric functions, ensure your calculator is in the correct mode (degrees or radians) depending on the problem.