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12.如图,已知点$A(2,0),B(0,4)$,且$∠1=∠2$,则$\tan∠OCA$的值为
2
。答案
12.2
13.[跨学科·物理]如图，CD是平面镜，光线从A出发经CD上点E反射后照射到B点.若入射角为α，AC⊥CD，BD⊥CD，垂足分别为点C，点D，且AC=3，BD=6，CD=12.求tan α的值.
答案
13.解:由光的反射定律可推出$∠ AEC=∠ BED.\because ∠ C=∠ D=90°,\therefore \mathrm{Rt}△ ACE∽\mathrm{Rt}△ BDE,\therefore \frac{AC}{BD}=\frac{CE}{DE},即\frac{3}{6}=\frac{CE}{12-CE},解得CE=4,\therefore \tan A=\frac{4}{3}.又\because α=∠ A,\therefore \tan α=\frac{4}{3}.$
14.如图,在等腰直角三角形ABC中,∠C=90°,点D在CB的延长线上,且BD=AB,求∠ADB的正切值. 
答案
14.解:设$AC=BC=x$,则$AB=\sqrt{2}x.\because BD=AB,\therefore BD=\sqrt{2}x$,$\therefore DC=BD+BC=\sqrt{2}x+x=(\sqrt{2}+1)x,\therefore \tan∠ ADB=\frac{AC}{CD}=\frac{x}{(\sqrt{2}+1)x}=\sqrt{2}-1.$
15.如图,在直角梯形ABCD中$,AB//CD,∠A=90°,AB=5,AD=4,S_{\mathrm{梯形}ABCD}=16,$试求tan B的值.
答案
15.解:过点$C$作$CE⊥ AB$于点$E$,则$CE=AD=4$.由梯形的面积$S_{\mathrm{梯形}ABCD}=16$,可得$\frac{1}{2}× (CD+5)× 4=16$,解得$CD=3$,$\therefore BE=AB-AE=AB-CD=5-3=2,\therefore \tan B=\frac{CE}{BE}=\frac{4}{2}=2.$
16.如图,在锐角△ABC中,AB=10 cm,BC=9 cm,△ABC的面积为27 cm²,求tan B的值.
答案
16.解:过点$A$作$AH⊥ BC$于点$H.\because S_{△ ABC}=\frac{1}{2}BC· AH=27$,$\therefore \frac{1}{2}× 9× AH=27,\therefore AH=6.\because AB=10,\therefore BH=\sqrt{AB^2-AH^2}=\sqrt{10^2-6^2}=8,\therefore \tan B=\frac{AH}{BH}=\frac{6}{8}=\frac{3}{4}.$
