Sure, let's walk through the solution.
We are given the sector area of turn B, which is [tex]\(\frac{51 \pi}{20}\)[/tex].
The formula for the area of a sector of a circle is:
[tex]\[
\text{Sector area} = \frac{\theta}{360} \times \pi r^2
\][/tex]
where [tex]\(\theta\)[/tex] is the central angle in degrees and [tex]\(r\)[/tex] is the radius of the circle. Assuming the radius [tex]\(r\)[/tex] is 1 (unit circle), the formula simplifies to:
[tex]\[
\text{Sector area} = \frac{\theta}{360} \times \pi
\][/tex]
Given the sector area, we can set up the equation:
[tex]\[
\frac{\theta}{360} \times \pi = \frac{51 \pi}{20}
\][/tex]
To solve for [tex]\(\theta\)[/tex], we first isolate it on one side of the equation.
1. Eliminate [tex]\(\pi\)[/tex] by dividing both sides by [tex]\(\pi\)[/tex]:
[tex]\[
\frac{\theta}{360} = \frac{51}{20}
\][/tex]
2. Next, solve for [tex]\(\theta\)[/tex] by multiplying both sides by 360:
[tex]\[
\theta = \frac{51}{20} \times 360
\][/tex]
3. Perform the multiplication:
[tex]\[
\theta = 51 \times 18
\][/tex]
[tex]\[
\theta = 918
\][/tex]
Thus, the measure of the central angle for turn B is:
[tex]\[
m \angle B = 918
\][/tex]
So the correct answer is:
[tex]\[
\boxed{918}
\][/tex]
