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To find the value of \( x \) in the given triangle, we can use the Law of Sines, which states:

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

where \( A, B, \) and \( C \) are the angles of the triangle, and \( a, b, \) and \( c \) are the sides opposite these angles, respectively.

Given:
- Angle \( \angle C = 55^\circ \)
- Side opposite \( x \) (which we will denote as \( a \)) is 3.5
- Side opposite \( 55^\circ \) (which we will denote as \( c \)) is 3.5
- The remaining side is 4

First, let's find the third angle \( \angle B \). Since the sum of the angles in a triangle is \( 180^\circ \):

\[
\angle B = 180^\circ - 55^\circ - x
\]

We can set up the Law of Sines as follows:

\[
\frac{\sin(x)}{3.5} = \frac{\sin(55^\circ)}{4}
\]

Solving for \( \sin(x) \):

\[
\sin(x) = \frac{3.5 \cdot \sin(55^\circ)}{4}
\]

Now, calculate \( \sin(55^\circ) \):

\[
\sin(55^\circ) \approx 0.8192
\]

Substitute this value back in:

\[
\sin(x) = \frac{3.5 \cdot 0.8192}{4}
\]
\[
\sin(x) \approx \frac{2.8672}{4}
\]
\[
\sin(x) \approx 0.7168
\]

Now, find \( x \):

\[
x = \sin^{-1}(0.7168)
\]

Using a calculator to find the inverse sine:

\[
x \approx 45.7^\circ
\]

Thus, the value of \( x \) is approximately \( 45.7^\circ \).

Would you like more details or have any questions?

Here are some additional questions you might find interesting:
1. How do you use the Law of Cosines in triangle problems?
2. Can you explain the difference between acute, obtuse, and right triangles?
3. What is the significance of the Pythagorean theorem in geometry?
4. How do you determine the area of a triangle given its sides?
5. How can you use trigonometric identities to simplify expressions?

**Tip:** Always ensure your calculator is set to the correct mode (degrees or radians) when solving trigonometric problems.

Learn how to find the value of angle x in a triangle using the Law of Sines. This problem involves trigonometric calculations and the application of the Law of Sines formula to solve for an unknown angle. Suitable for advanced high school students.

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