Use the stated variant of the Law of Cosines, b2=a2+c2−2ac⋅cosB , to find the value of side b in a right triangle, where a=3 , c=5 , and B=53.13 . Round your answer to the nearest whole number.

Answer 1

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:

1. Calculate \( a^2 \) and \( c^2 \):

\[ a^2 = 3^2 = 9 \] \[ c^2 = 5^2 = 25 \]

2. Now substitute these values into the formula:

\[ b^2 = 9 + 25 - 2 \cdot 3 \cdot 5 \cdot 0.6 \]

3. Calculate \( 2 \cdot 3 \cdot 5 \cdot 0.6 \):

\[ 2 \cdot 3 \cdot 5 \cdot 0.6 = 6 \cdot 5 \cdot 0.6 = 30 \cdot 0.6 = 18 \]

4. Substitute back:

\[ b^2 = 9 + 25 - 18 \] \[ b^2 = 34 - 18 = 16 \]

5. Take the square root to find \( b \):

\[ b = \sqrt{16} = 4 \]

Thus, the value of side \( b \) is approximately:

\[ \boxed{4} \]