已知三角形ABC的三邊長為a=2√3,b=2√2,c=√6+√2,求三角形ABC的各角度數(shù)
cosC=(a²+b²-c²)/2ab=[12+8-(6+2√12+2)]/(8√6)=(12-2√12)/(8√6)=(12-4√3)/(8√6)=(3-√3)/(2√6)
=(3√6-3√2)/12=(√6-√2)/4,故C=75º.
cosA=(b²+c²-a²)/2bc=[8+(8+4√3)-12]/[4√2(√6+√2)]=(4+4√3)/[4√2(√6+√2)]=(1+√3)/[√2(√6+√2)
=(1+√3)(√6-√2)/(4√2)=(2√2)/(4√2)=1/2,故A=60º；
B=180º-(75º+60º)=45º