Answer:
20.9 units
Explanation:
To find side b in triangle ABC, we can use the Law of Cosines, which states:
\[ c^2 = a^2 + b^2 - 2ab \cdot \cos(B) \]
Given that \( B = 92° \), \( a = 38 \), and \( c = 28 \), let's calculate b:
\[ 28^2 = 38^2 + b^2 - 2 \cdot 38 \cdot b \cdot \cos(92°) \]
Now, calculate the value inside the cosine function:
\[ \cos(92°) \approx -0.139 \]
\[ 28^2 = 38^2 + b^2 - 2 \cdot 38 \cdot b \cdot (-0.139) \]
\[ 784 = 1444 + b^2 + 10.564b \]
Rearrange the equation:
\[ b^2 + 10.564b - 660 = 0 \]
Now, solve for b using the quadratic formula:
\[ b = \frac{-10.564 \pm \sqrt{10.564^2 - 4 \cdot 1 \cdot (-660)}}{2 \cdot 1} \]
\[ b = \frac{-10.564 \pm \sqrt{111.408496 + 2640}}{2} \]
\[ b = \frac{-10.564 \pm \sqrt{2751.408496}}{2} \]
\[ b = \frac{-10.564 \pm 52.46}{2} \]
Now, we'll have two possible solutions for b:
1. \( b = \frac{-10.564 + 52.46}{2} = \frac{41.896}{2} = 20.948 \)
2. \( b = \frac{-10.564 - 52.46}{2} = \frac{-63.024}{2} = -31.512 \)
Since we're dealing with a triangle, the length of a side cannot be negative. Therefore, we discard the negative solution.
So, side b is approximately 20.9 units.