第97页 - 通城学典课时作业本九年级数学苏科版江苏专用

解：$(1) $∵$ $四边形$ ABCD $是正方形，

∴$ AD = DC，$$∠ADC = 90°。$

∵$ CF\perp DE，$

∴$ ∠CFD=∠CFG = 90°。$

∵$ AG// CF，$

∴$ ∠AGD=∠CFG = 90°。$

∴$ ∠AGD=∠CFD。$

又 ∵$ ∠ADG+∠CDE=∠ADC = 90°，$

在$ Rt\triangle CFD $中，$∠DCF+∠CDE = 90°，$

∴$ ∠ADG=∠DCF。$

在$ \triangle ADG $和$ \triangle DCF $中，

$\begin {cases}∠AGD=∠DFC\\∠ADG=∠DCF\\AD = DC\end {cases}，$

∴$ \triangle ADG\cong \triangle DCF(\mathrm {AAS})。$

$(2) $设正方形$ ABCD $的边长为$ 2a，$

∵$ E $是$ AB $的中点，

∴$ AE=\frac {1}{2}×2a = a。$

∴$ $在$ Rt\triangle DAE $中，$DE=\sqrt {AD^2+AE^2}=\sqrt {(2a)^2+a^2}=\sqrt {5}a。$

∴$ \sin ∠ADE=\frac {AE}{DE}=\frac {a}{\sqrt {5}a}=\frac {\sqrt {5}}{5}。$

由$(1)，$得$ ∠ADE=∠DCF = a，$

∴$ \sin a=\frac {\sqrt {5}}{5}。$

解：过点$D$作$DE // AC$交$BC$于点$E，$则$∠CDE = ∠ACD = 90^\circ 。$

设$DE = x，$

因为$\tan ∠BCD = \frac {DE}{CD} = \frac {1}{3}，$

所以$CD = 3x。$

由于$D$是$AB$的中点且$DE // AC，$

所以$DE$是$\triangle ABC$的中位线，故$AC = 2DE = 2x。$

在$\text{Rt}\triangle ACD$中，$AD = \sqrt {AC^2 + CD^2} = \sqrt {(2x)^2 + (3x)^2} = \sqrt {13}x。$

因此，$\sin A = \frac {CD}{AD} = \frac {3x}{\sqrt {13}x} = \frac {3\sqrt {13}}{13}，$$\cos A = \frac {AC}{AD} = \frac {2x}{\sqrt {13}x} = \frac {2\sqrt {13}}{13}，$

$\tan A = \frac {CD}{AC} = \frac {3x}{2x} = \frac {3}{2}。$

解：$(1)$过点$O$作$OD \perp AB$于点$D。$

因为$OD \perp AB$且$O$是圆心，$AB = 8，$

所以$BD = \frac {1}{2}AB = 4。$

在$\text{Rt}\triangle ODB$中，$\cos ∠ABC = \frac {BD}{OB} = \frac {4}{5}，$则$OB = \frac {BD}{\frac {4}{5}} = 5，$

即$\odot O$的半径为$5。$

$ (2)$过点$C$作$CE \perp AB$于点$E。$

因为$OC = \frac {1}{2}OB，$$OB = 5，$

所以$OC = \frac {5}{2}，$$BC = OB + OC = 5 + \frac {5}{2} = \frac {15}{2}。$

因为$OD \perp AB，$$CE \perp AB，$

所以$OD // CE，$则$\frac {OB}{BC} = \frac {BD}{BE}，$即$\frac {5}{\frac {15}{2}} = \frac {4}{BE}，$解得$BE = 6。$

所以$AE = AB - BE = 8 - 6 = 2。$

在$\text{Rt}\triangle BCE$中，$CE = \sqrt {BC^2 - BE^2} = \sqrt {(\frac {15}{2})^2 - 6^2} = \frac {9}{2}。$

因此，$\tan ∠BAC = \frac {CE}{AE} = \frac {\frac {9}{2}}{2} = \frac {9}{4}。$