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To solve for side \( b \) in this triangle, we can use the Law of Cosines. The triangle is defined by:

- \( AC = 4 \)
- \( BC = 2 \)
- Angle \( C = 55^\circ \)

The Law of Cosines states:

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

Substituting the known values:

\[
b^2 = 4^2 + 2^2 - 2 \cdot 4 \cdot 2 \cdot \cos(55^\circ)
\]

Calculating each part:

1. \( 4^2 = 16 \)
2. \( 2^2 = 4 \)
3. \( 2 \cdot 4 \cdot 2 = 16 \)
4. \( \cos(55^\circ) \approx 0.5736 \)

So,

\[
b^2 = 16 + 4 - 16 \cdot 0.5736
\]
\[
b^2 = 20 - 9.1776
\]
\[
b^2 \approx 10.8224
\]
\[
b \approx \sqrt{10.8224} \approx 3.29
\]

Thus, \( b \approx 3.29 \) (rounded to two decimal places).

Would you like further clarification on any steps?

---

1. How do we use the Law of Sines to solve for other angles in this triangle?
2. Can we solve for all angles in a triangle given only sides?
3. What is the Law of Cosines useful for beyond this problem?
4. How does rounding affect the final answer in trigonometric calculations?
5. Why might we choose the Law of Cosines over the Pythagorean Theorem here?

**Tip:** Always check if a triangle is right before using trigonometric laws; right triangles simplify to the Pythagorean Theorem.

Solve the triangle given side AC = 4, side BC = 2, and angle C = 55°, and find side b to two decimal places.

Learn how to solve for the unknown side b in a triangle with side lengths AC = 4, BC = 2, and angle C = 55°. This problem demonstrates how to use the Law of Cosines, with a detailed solution and step-by-step calculations. Suitable for high school trigonometry students in Grades 10-12.

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