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第59页

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$3038\pi$

$\frac{5\pi}{4}-\frac{3}{2}$

 解：（1）直线$BD$与$\odot O$相切.理由如下：连接$OB.$因为$\angle ACB = 60^{\circ}，$所以$\angle AOB = 2\angle ACB = 120^{\circ}，$所以$\angle BOD = 180^{\circ}-\angle AOB = 60^{\circ}.$因为$\angle ADB = 30^{\circ}，$所以$\angle OBD = 180^{\circ}-\angle BOD - \angle ADB = 90^{\circ}，$所以$OB\perp BD.$因为$OB$是$\odot O$的半径，所以直线$BD$与$\odot O$相切.

 （2）因为$\angle BOD = 60^{\circ}，$所以$\angle BAD=\frac{1}{2}\angle BOD = 30^{\circ}.$因为$\angle ADB = 30^{\circ}，$所以$\angle BAD = \angle ADB，$所以$BD = AB = 4\sqrt{3}.$因为$\angle OBD = 90^{\circ}，$所以$OB=\frac{1}{2}OD.$设$OB = x，$则$OD = 2x，$所以$BD=\sqrt{OD^{2}-OB^{2}}=\sqrt{3}x，$所以$\sqrt{3}x = 4\sqrt{3}，$解得$x = 4，$所以$OB = 4，$所以$S_{\triangle OBD}=\frac{1}{2}OB\cdot BD = 8\sqrt{3}，$$S_{扇形OBE}=\frac{60\pi\times4^{2}}{360}=\frac{8\pi}{3}，$所以$S_{阴影}=S_{\triangle OBD}-S_{扇形OBE}=8\sqrt{3}-\frac{8\pi}{3}.$

 解：（1）如图，取$AD$的中点$E，$连接$OE.$因为$OA\perp OB，$所以$\angle AOB = 90^{\circ}，$所以$OE = AE = DE=\frac{1}{2}AD.$因为$CD=\frac{1}{2}AD，$所以$AE = CD.$因为$OA = OC，$所以$\angle OAE = \angle OCD.$在$\triangle OAE$和$\triangle OCD$中，$\begin{cases}OA = OC\\\angle OAE = \angle OCD\\AE = CD\end{cases}，$所以$\triangle OAE\cong\triangle OCD，$所以$\angle AOE = \angle COD，$$OE = OD，$所以$DE = OE = OD，$所以$\triangle ODE$是等边三角形，所以$\angle DOE = 60^{\circ}，$所以$\angle COD = \angle AOE = \angle AOB - \angle DOE = 30^{\circ}，$即$\angle BOC = 30^{\circ}.$

 （2）如图，过点$O$作$OF\perp AC$于点$F，$则$\angle OFA = 90^{\circ}，$$AF=\frac{1}{2}AC.$因为$\angle AOB = 90^{\circ}，$$\angle BOC = 30^{\circ}，$所以$\angle AOC = \angle AOB + \angle BOC = 120^{\circ}.$因为$OA = OC，$所以$\angle OAC = \angle OCA=\frac{1}{2}(180^{\circ}-\angle AOC)=30^{\circ}，$所以$OF=\frac{1}{2}OA.$因为$OA = 2，$所以$OF = 1，$所以$AF=\sqrt{OA^{2}-OF^{2}}=\sqrt{3}，$所以$AC = 2AF = 2\sqrt{3}.$因为$CD=\frac{1}{2}AD，$所以$CD=\frac{1}{3}AC=\frac{2\sqrt{3}}{3}，$所以$S_{\triangle OCD}=\frac{1}{2}CD\cdot OF=\frac{\sqrt{3}}{3}.$因为$OB = OA = 2，$所以$S_{扇形OBC}=\frac{30\pi\times2^{2}}{360}=\frac{\pi}{3}，$所以线段$BD$、线段$CD$和$\overset{\frown}{BC}$围成的图形的面积$S = S_{扇形OBC}-S_{\triangle OCD}=\frac{\pi - \sqrt{3}}{3}.$