△ABC 中,AB =6,AD =4, CD=5, BD=4,則 BC=【 】。?
回答 (2)
Sol
加上D為BC上一點,BC改為AC
x=AC
Cos(∠ADB)+Cos(∠ADC)=0
(16+16-36)/(2*4*4)+(16+25-x^2)/(2*4*5)=0
-1/8+(41-x^2)/40=0
-5+41-x^2=0
x^2=36
x=6
設D點在三角形之內,並令到AB =6,AD =4, CD=5, BD=4
BD=AD=4 (已知)
CD=CD=5 (已知)
AC=BC
ADC=BDC (全等三角形S.S.S)
AB^2=AD^2+DB^2-2(AD)(DB)cos(∠ADB)
6^2=4^2+4^2-2(4)(4)cos(∠ADB)
∠ADB=97.18075578
∠BDC=(180-97.18075578)/2
∠BDC=41.40962211
BC^2=BD^2+DC^2-2(BD)(DC)cos(∠BDC)
BC^2=4^2+5^2-2(4)(5)cos(41.40962211)
BC=3.31662479
收錄日期: 2021-04-18 15:08:23
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