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電子課本網 第61頁

第61頁

信息發(fā)布者:
解:設?$AE=x,$?則?$BE=3x$?
∵四邊形?$ABCD$?是正方形,?$M$?是?$AD$?的中點
∴?$BC=CD=4x,$??$AM=2x$?
∴?$EC=\sqrt{(3x)^2+(4x)^2}=5x,$??$EM=\sqrt{x^2+(2x)^2}=\sqrt 5x,$
??$CM=\sqrt{(2x)^2+(4x)^2}=2\sqrt 5x$?
∴?$EM^2+CM^2=CE^2$?
∴?$△CEM$?是直角三角形
∴?$sin ∠ECM=\frac {EM}{CE}=\frac {\sqrt 5x}{5x}=\frac {\sqrt 5}5$?
解:∵?$s in A+s in B=\frac {4}{3}$?
∴?$(s in A+ s in B)^2=\frac {16}{9}$?
∴?$s in ^2\ \mathrm {A}+ s in ^2B+2\ \mathrm {s} in A · s in B=\frac {16}{9}$?
∵?$s in B= cos A$?
∴?$s in ^2\ \mathrm {A}+ cos ^2\ \mathrm {A}+2\ \mathrm {s} in A · s in B=\frac {16}{9}$?
∴?$2\ \mathrm {s} in A · s in B=\frac {7}{9}$?
∴?$(s in A-s in B)^2=1- \frac {7}{9}=\frac {2}{9}$?
∴?$s in A- s in B=±\frac {\sqrt 2}3$?
?$(1)$?證明:∵?$AC=CD$?
∴?$\widehat{AC}=\widehat{CD}$?
∴?$∠CAD=∠ABC$?
∵?$AB$?是直徑
∴?$∠ACB=90°$?
∴?$∠ABC+∠BAC=90°$?
∵?$AE=AF,$??$∠ACB=90°$?
∴?$∠FAC=∠CAD,$??$CE=FC$?
∴?$∠FAC=∠CAD=∠ABC$?
∴?$∠FAC+∠CAB=90°$?
∴?$AF⊥AB$?
又∵?$AB$?是直徑
∴?$AF $?是?$⊙O$?的切線
?$ (2) $?連接?$BD$?
∵?$cos ∠ABC=cos∠CAD=\frac {AC}{AE}=\frac {4}{5},$??$AC=4$?
∴?$AE=5$?
∴?$AF=AE=5,$??$CE=\sqrt{AE^2-AC^2}=\sqrt{5^2-4^2}=3$?
∴?$CE=CF=3$?
∵?$tan F = \frac {AC}{CF}= \frac {AB}{AF}$?
∴?$ \frac {4}{3} =\frac {AB}{5}$?
∴?$AB=\frac {20}{3}$?
∵?$cos∠ABC=\frac {4}{5}=\frac {AB}{BF}$?
∴?$BF=\frac {25}{3}$?
∴?$ BE=BF-EF=\frac {25}{3} -(3+3)=\frac {7}{3}$?
∵?$∠CAD=∠CBD,$??$∠AEC=∠BED$?
∴?$△ACE∽△BDE$?
∴?$\frac {AE}{BE}=\frac {EC}{ED} $?
∴?$\frac {5}{\frac {7}{3}}= \frac {3}{ED} $?
∴?$ED=\frac {7}{5}$?
∴?$AD=ED+AE=\frac {7}{5} +5=\frac {32}{5}$?
∴?$cos ∠BAD= \frac {AD}{AB} =\frac {\frac {32}{5}}{\frac {20}{3}} =\frac {24}{25}$?