第47页

信息发布者:
130或100或160
$\frac{10}{3}$或10
45°或22.5°或67.5°
6
84°,117°,124°,103.5°,126°
解:(4)如图⑦,当$DA = DE$时,又$AD = DC,$$BE = DE,$$\therefore\angle CAD=\angle C = 27^{\circ}。$设$\angle B = x,$则$\angle BDE = x,$$\angle BAD=\angle AED = 2x。$$\because\angle B+\angle C+\angle BAC = 180^{\circ},$$\therefore x + 27^{\circ}+2x + 27^{\circ}=180^{\circ},$解得$x = 42^{\circ},$$\therefore\angle B = 42^{\circ}。$如图⑧,当$AD = AE$时,$\because AD = CD,$$BE = DE,$$\therefore\angle CAD=\angle C = 27^{\circ}。$设$\angle B = x,$则$\angle BDE = x,$$\angle ADE=\angle AED = 2x。$又$\angle ADB=\angle DAC+\angle C = 54^{\circ},$$\angle ADB=\angle ADE+\angle BDE = 3x,$$\therefore 3x = 54^{\circ},$解得$x = 18^{\circ},$$\therefore\angle B = 18^{\circ}。$如图⑨,当$AE = DE$时,又$BE = DE,$$\therefore\angle ADB = 90^{\circ}。$又$AD = CD,$$\therefore\angle C=\angle CAD=\frac{1}{2}\angle ADB = 45^{\circ}\neq27^{\circ}。$故$\angle B$的度数为18°或42°。