四边形ABCD中,∠A=∠CBD,AB=4cm,AD=6cm,BD=5cm,BC=7.5cm,求CD的长

由余弦定理,得
cos∠A=(AB^2+AD^2-BD^2) / (2*AB*AD)
=(4^2+6^2-5^2) / (2*4*6)
=(16+36-25) / 48
=27/48
=9/16
CD^2=BC^2+BD^2-2*BC*BD*cos∠CBD
=7.5^2+5^2-2*7.5*5*cos∠A
=56.25+25-75*9/16
=1975/16
所以
CD=√(1975/16)=5√79/4(cm)我还没教余弦由余弦定理，得cos∠A=(AB^2+AD^2-BD^2) / (2*AB*AD) =(4^2+6^2-5^2) / (2*4*6) =(16+36-25) / 48 =27/48 =9/16CD^2=BC^2+BD^2-2*BC*BD*cos∠CBD=7.5^2+5^2-2*7.5*5*cos∠A=56.25+25-75*9/16=625/16所以CD=√(625/16)=25/4=6.25(cm) 没学怎么会出这样的题？我教的是相似三角形∵∠A=∠CBD，AB/BD=4/5，AD/BC=6/7.5=4/5∴AB/BD=AD/BC∴△ABD∽△BDC∴BD/DC=4/5即5/DC=4/5∴DC=5*5/4=25/4=6.25(cm)