To find the value of side \( b \) in a right triangle using the Law of Cosines, we can use the specified formula:
\[ b^2 = a^2 + c^2 - 2ac \cdot \cos B \]
Given:
- \( a = 3 \)
- \( c = 5 \)
- \( B = 53.13^\circ \)
First, we will find \( \cos(53.13^\circ) \).
Using a calculator, we find:
\[ \cos(53.13^\circ) \approx 0.6 \]
Now we can substitute the values into the Law of Cosines equation:
- Calculate \( a^2 \) and \( c^2 \):
\[ a^2 = 3^2 = 9 \] \[ c^2 = 5^2 = 25 \]
- Now substitute these values into the formula:
\[ b^2 = 9 + 25 - 2 \cdot 3 \cdot 5 \cdot 0.6 \]
- Calculate \( 2 \cdot 3 \cdot 5 \cdot 0.6 \):
\[ 2 \cdot 3 \cdot 5 \cdot 0.6 = 6 \cdot 5 \cdot 0.6 = 30 \cdot 0.6 = 18 \]
- Substitute back:
\[ b^2 = 9 + 25 - 18 \] \[ b^2 = 34 - 18 = 16 \]
- Take the square root to find \( b \):
\[ b = \sqrt{16} = 4 \]
Thus, the value of side \( b \) is approximately:
\[ \boxed{4} \]