Question

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:

\[
\boxed{4}
\]