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17. (广元中考)如图，在直角坐标系中，点A的坐标为(1，0)，点B的坐标为(0，－3)，点C在x轴上，且点C在点A右边，连接AB，BC. 若tan ∠ABC = $\frac{1}{3}$，则点C的坐标为________.
答案：
$(\frac{9}{4},0)$
18. (齐齐哈尔中考)在△ABC中，AB = 3$\sqrt{6}$，AC = 6，∠B = 45°，则BC = ________.
答案：
$3\sqrt{3}+3$或$3\sqrt{3}-3$
19. (6分)计算：
(1) $\frac{\sin 60^{\circ}-1}{\tan 60^{\circ}-2\tan 45^{\circ}}-\sqrt{3}\cos 30^{\circ}+\sqrt{2}\sin 45^{\circ}$； (2) $\frac{\sqrt{2}}{4}\sin 45^{\circ}+\cos ^{2}30^{\circ}-\frac{1}{2\tan 60^{\circ}}+2\sin 60^{\circ}$.
(1) $\frac{\sin 60^{\circ}-1}{\tan 60^{\circ}-2\tan 45^{\circ}}-\sqrt{3}\cos 30^{\circ}+\sqrt{2}\sin 45^{\circ}$； (2) $\frac{\sqrt{2}}{4}\sin 45^{\circ}+\cos ^{2}30^{\circ}-\frac{1}{2\tan 60^{\circ}}+2\sin 60^{\circ}$.
答案：
(1) 原式$=\frac{\frac{\sqrt{3}}{2}-1}{\sqrt{3}-2\times1}-\sqrt{3}\times\frac{\sqrt{3}}{2}+\sqrt{2}\times\frac{\sqrt{2}}{2}=\frac{1}{2}-\frac{3}{2}+1 = 0$
(2) 原式$=\frac{\sqrt{2}}{4}\times\frac{\sqrt{2}}{2}+(\frac{\sqrt{3}}{2})^2-\frac{1}{2\times\sqrt{3}}+2\times\frac{\sqrt{3}}{2}=\frac{1}{4}+\frac{3}{4}-\frac{\sqrt{3}}{6}+\sqrt{3}=1+\frac{5\sqrt{3}}{6}$
(1) 原式$=\frac{\frac{\sqrt{3}}{2}-1}{\sqrt{3}-2\times1}-\sqrt{3}\times\frac{\sqrt{3}}{2}+\sqrt{2}\times\frac{\sqrt{2}}{2}=\frac{1}{2}-\frac{3}{2}+1 = 0$
(2) 原式$=\frac{\sqrt{2}}{4}\times\frac{\sqrt{2}}{2}+(\frac{\sqrt{3}}{2})^2-\frac{1}{2\times\sqrt{3}}+2\times\frac{\sqrt{3}}{2}=\frac{1}{4}+\frac{3}{4}-\frac{\sqrt{3}}{6}+\sqrt{3}=1+\frac{5\sqrt{3}}{6}$
20. (8分)如图，AD是△ABC的中线，tan B = $\frac{1}{5}$，cos C = $\frac{\sqrt{2}}{2}$，AC = $\sqrt{2}$. 求：
(1) BC的长；
(2) ∠ADC的正弦值.
(1) BC的长；
(2) ∠ADC的正弦值.
答案：
(1) 如图,过点$A$作$AH\perp BC$于点$H$. 在$Rt\triangle ACH$中,$\because\cos C=\frac{CH}{AC}=\frac{\sqrt{2}}{2},AC = \sqrt{2},\therefore CH = 1$. $\therefore$在$Rt\triangle ACH$中,由勾股定理,得$AH=\sqrt{AC^{2}-CH^{2}}=\sqrt{(\sqrt{2})^{2}-1^{2}} = 1$. 在$Rt\triangle ABH$中,$\because\tan B=\frac{AH}{BH}=\frac{1}{5},\therefore BH = 5$. $\therefore BC = BH + CH = 5 + 1 = 6$
(2) $\because AD$是$\triangle ABC$的中线,$\therefore BD = CD$. $\because BC = 6,CH = 1,\therefore CD = 3,DH = 2$. 在$Rt\triangle ADH$中,由勾股定理,得$AD=\sqrt{AH^{2}+DH^{2}}=\sqrt{1^{2}+2^{2}}=\sqrt{5}$. $\therefore\sin\angle ADH=\frac{AH}{AD}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$. $\therefore\angle ADC$的正弦值为$\frac{\sqrt{5}}{5}$
(1) 如图,过点$A$作$AH\perp BC$于点$H$. 在$Rt\triangle ACH$中,$\because\cos C=\frac{CH}{AC}=\frac{\sqrt{2}}{2},AC = \sqrt{2},\therefore CH = 1$. $\therefore$在$Rt\triangle ACH$中,由勾股定理,得$AH=\sqrt{AC^{2}-CH^{2}}=\sqrt{(\sqrt{2})^{2}-1^{2}} = 1$. 在$Rt\triangle ABH$中,$\because\tan B=\frac{AH}{BH}=\frac{1}{5},\therefore BH = 5$. $\therefore BC = BH + CH = 5 + 1 = 6$
(2) $\because AD$是$\triangle ABC$的中线,$\therefore BD = CD$. $\because BC = 6,CH = 1,\therefore CD = 3,DH = 2$. 在$Rt\triangle ADH$中,由勾股定理,得$AD=\sqrt{AH^{2}+DH^{2}}=\sqrt{1^{2}+2^{2}}=\sqrt{5}$. $\therefore\sin\angle ADH=\frac{AH}{AD}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$. $\therefore\angle ADC$的正弦值为$\frac{\sqrt{5}}{5}$
21. (10分)(安徽中考)科技社团选择在学校游泳池进行一次光的折射实验，如图，光线自点B处发出，经水面点E折射到池底点A处. 已知BE与水平线的夹角α = 36.9°，点B到水面的距离BC = 1.20 m，点A处的水深为1.20 m，到池壁的水平距离AD = 2.50 m. 点B，C，D在同一条竖直线上，所有点都在同一竖直平面内. 记入射角为β，折射角为γ，求$\frac{\sin \beta}{\sin \gamma}$的值(结果取小数点后一位，参考数据：sin 36.9°≈0.60，cos 36.9°≈0.80，tan 36.9°≈0.75).
答案：
过点$E$作$EH\perp AD$于点$H$, 易得四边形$CDHE$是矩形,$\therefore EH = CD = 1.20\ m,DH = CE$. 由题意,得$\angle CEB=\alpha = 36.9^{\circ},\angle CBE=\beta,\therefore CE=\frac{BC}{\tan36.9^{\circ}}\approx\frac{1.20}{0.75}=1.60(m),AH = AD - DH = AD - CE = 2.50 - 1.60 = 0.90(m)$. $\therefore$在$Rt\triangle AEH$中,由勾股定理,得$AE=\sqrt{AH^{2}+EH^{2}}=\sqrt{0.90^{2}+1.20^{2}} = 1.50(m)$. $\therefore\sin\gamma=\frac{AH}{AE}=\frac{0.90}{1.50}=0.60$. $\because\sin\beta=\sin\angle CBE=\frac{CE}{BE}=\cos\angle CEB=\cos\alpha\approx0.80,\therefore\frac{\sin\beta}{\sin\gamma}=\frac{0.80}{0.60}\approx1.3$
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