一个半径为 $r$ 的圆形蛋糕被切成 22 个扇形。这些扇形的面积构成一个递增的等差数列。第八个扇形的面积是最小扇形面积的两倍。用 $\pi$ 表示最大扇形的角。 [扇形面积 = $\tfrac{1}{2} r^2 \theta$，其中 $\theta$ 是扇形的角]

1. Establish a relationship between the area of the smallest sector and the common difference.

Let the area of the smallest sector be $a$ and the common difference of the arithmetic progression be $d$.
Given that the area of the eighth sector is twice the area of the smallest sector:
$A_8 = 2A_1$
$a + 7d = 2a$
$a = 7d$

2. Set up an equation for the sum of all 22 sectors and solve for $d$.

The sum of the areas of all 22 sectors is the total area of the circular cake, which is $\pi r^2$.
$S_{22} = \frac{22}{2}[2a + (22 - 1)d] = \pi r^2$
$11[2(7d) + 21d] = \pi r^2$
$11[14d + 21d] = \pi r^2$
$11[35d] = \pi r^2$
$385d = \pi r^2$
$d = \frac{\pi r^2}{385}$

3. Calculate the area of the largest sector ($A_{22}$).

The largest sector is the 22nd sector.
$A_{22} = a + 21d$
$A_{22} = 7d + 21d = 28d$
$A_{22} = 28 \left( \frac{\pi r^2}{385} \right)$
$A_{22} = \frac{4}{55} \pi r^2$

4. Find the angle of the largest sector.

Let the angle of the largest sector be $\theta$.
Using the area of a sector formula:
$\frac{1}{2} r^2 \theta = \frac{4}{55} \pi r^2$
$\frac{1}{2} \theta = \frac{4}{55} \pi$
$\theta = \frac{8\pi}{55}$