《2025年名师学案九年级数学下册人教版》

1.(2024·云南改编)在Rt△ABC中,∠B = 90°,AB = 3,BC = 4,则sin C等于( )
A.$\frac{3}{4}$ B.$\frac{3}{5}$ C.$\frac{4}{5}$ D.$\frac{4}{3}$

答案： B

2.如图,PA与⊙O相切于点A,PC经过⊙O的圆心且与该圆相交于B,C两点.若OB = 3,PB = 2,则cos P = ________.

答案： $\frac{4}{5}$

3.如图,在Rt△ABC中,∠B = 90°,AB = 5,BC = 12,将△ABC绕点A逆时针旋转得到△ADE,使得点D落在AC上,则tan∠ECD的值为________.

答案： $\frac{3}{2}$

4.在△ABC中,∠C = 90°,sin B = $\frac{4}{5}$,则tan A等于( )
A.$\frac{4}{3}$ B.$\frac{3}{4}$ C.$\frac{3}{5}$ D.$\frac{4}{5}$

答案： B

5.如图,在菱形ABCD中,DE⊥AB于点E,cos A = $\frac{3}{5}$,则tan∠DBE的值是( )

A.$\frac{1}{2}$ B.2 C.$\frac{3}{5}$ D.$\frac{4}{3}$

答案： B

6.如图,在△ABC中,D是AB的中点,DC⊥AC,且tan∠BCD = $\frac{1}{3}$,则sinA的值是________.

【点拨】过点D作DE//AC交BC于E,可得∠CDE = 90°.设DE = x,则CD = 3x.利用△BDE∽△BAC可得AC = 2DE = 2x,在Rt△ACD中利用勾股定理和正弦定义解答.

答案： $\frac{3\sqrt{13}}{13}$

7.如图,在正方形ABCD中,M是AD的中点,AE∶BE = 1∶3.求cos∠ECM的值.

答案： 解:设$AE = x,BE = 3x$,$\because$正方形$ABCD$,$\therefore AB = BC = CD = AD = 4x$,$\angle A=\angle B=\angle D = 90^{\circ}$.$\because M$是$AD$的中点,$\therefore AM = DM=\frac{1}{2}AD = 2x$.$\therefore EC=\sqrt{BE^{2}+CB^{2}} = 5x$,$EM=\sqrt{AE^{2}+AM^{2}}=\sqrt{5}x$,$CM=\sqrt{DM^{2}+DC^{2}} = 2\sqrt{5}x$.$\therefore EM^{2}+CM^{2}=(\sqrt{5}x)^{2}+(2\sqrt{5}x)^{2}=25x^{2}$.$\because EC^{2}=(5x)^{2}=25x^{2}$,$\therefore EM^{2}+CM^{2}=EC^{2}$.$\therefore \angle EMC = 90^{\circ}$.$\therefore \cos\angle ECM=\frac{CM}{EC}=\frac{2\sqrt{5}x}{5x}=\frac{2}{5}\sqrt{5}$.