第93页 - 苏科版九年级数学课课练答案（上下册）

45°

60°

80

$\frac {\sqrt{3}}{2}$

$\frac {\sqrt{6}}{6}$

$解：原式=(\frac {1}{2})²+(\frac {\sqrt{3}}{2})²-\sqrt{2}×\frac {\sqrt{2}}{2}$

$=1-1$

$=0$

$解：原式=\frac {\sqrt{3}}{3}+1-2×\frac {1}{2}+\sqrt{3}$

$=\frac {\sqrt{3}}{3}+1-1+\sqrt{3}$

$=\frac {4\sqrt{3}}{3}$

$解：原式=\frac {\frac {\sqrt{2}}{2}}{\frac {\sqrt{3}}{2}+1}-\frac {\frac {\sqrt{2}}{2}}{1-\frac {\sqrt{3}}{2}}$

$=\frac {\frac {\sqrt{2}×(1-\frac {\sqrt{3}}{2})}{2}}{\frac {1}{4}}-\frac {\frac {\sqrt{2}}{2}×(1-\frac {\sqrt{3}}{2})}{\frac {1}{4}}$

$=2\sqrt{2}×[(1-\frac {\sqrt{3}}{2})-(1+\frac {\sqrt{3}}{2})]$

$=2\sqrt{2}×(-\sqrt{3})$

$=-2\sqrt{6}$

$解：原式=2×\frac {1}{2}+\frac {1}{3}+1-2$

$=1+\frac {1}{3}-1$

$=\frac {1}{3}$

$解： (1)在Rt△ABC中$

$因为∠A=30°，∠C=90°，∠B=60°$

$因为a=4\sqrt{3}$

$所以b=\frac {a}{tan_{30}°}=12，c=\frac {a}{sin_{30}°}=8\sqrt{3}$

$(2)在Rt△ABC中$

$因为a= 6\sqrt{2}，c=12，∠C = 90°$

$所以b=\sqrt{c²-a²}= 6\sqrt{2}$

$所以sinA=\frac {a}{c}=\frac {\sqrt{2}}{2}，sinB=\frac {b}{c}=\frac {\sqrt{2}}{2}$

$所以∠A=45°，∠B=45°$

解:过点$A$作$AE⊥BD，$交$BD$的延长线于点$E ，$如图所示

$因为DB⊥BC， AE⊥DB$

$所以AE//BC，∠AED=∠DBC= 90°$

$所以∠EAD=∠C=30°， $

$所以△ADE∽△CDB$

$因为\frac {AD}{DC}=\frac {1}{2}$

$所以\frac {AE}{BC}=\frac {DE}{DB}=\frac {1}{2}$

$设DE=x ，则DB=2x， BE=3x$

$在Rt△ADE中$

$因为∠EAD=30°， DE=x$

$所以AE=\sqrt{3}x$

$所以tan∠ABD=\frac {AE}{BE}=\frac {\sqrt{3}x}{3x}=\frac {\sqrt{3}}{3}$