O 是穿过 B、E、D 和 C 的圆的圆心。角 $DOC = 84^\circ$ 且角 $CBO = 54^\circ$。$PQ$ 是在点 D 处与圆相切的切线。 求以下各角的值，并给出每一步的理由： 角 BAE， 角 QDC； 角 ECB， 角 CBE。 图中显示了 3 个圆，它们的直径分别为 d cm、2d cm 和 3d cm。请用 π 和 d 表示非阴影区域与阴影区域的面积之差。

1. Find the angle between radius $OD$ and chord $DC$.

Since $OD=OC$ radii of the circle, $\triangle DOC$ is isosceles.

$\angle ODC=\angle OCD=\frac{180^\circ-84^\circ}{2}=48^\circ$

2. Use the tangent-radius property to find $\angle QDC$.

Since $PQ$ is a tangent at $D$, $OD\perp QD$.

Hence
$\angle QDC=90^\circ-\angle ODC=90^\circ-48^\circ=42^\circ.$

1. Write the radii of the three circles from their diameters.

The diameters are $d$, $2d$ and $3d$.

So the radii are
$\frac d2,\quad d,\quad \frac{3d}{2}.$

2. Find the total unshaded area.

Unshaded area = area of inner circle + area of outer annulus.

$\pi\left(\frac d2\right)^2+\left[\pi\left(\frac{3d}{2}\right)^2-\pi d^2\right]$

$=\frac14\pi d^2+\left(\frac94\pi d^2-\pi d^2\right)$

$=\frac14\pi d^2+\frac54\pi d^2$

$=\frac32\pi d^2$

3. Find the shaded area and subtract it from the unshaded area.

Shaded area = middle circle area $-$ inner circle area.

$\pi d^2-\pi\left(\frac d2\right)^2$

$=\pi d^2-\frac14\pi d^2$

$=\frac34\pi d^2$

Difference between unshaded and shaded area:

$\frac32\pi d^2-\frac34\pi d^2=\frac34\pi d^2$