1.已知△ABC中 ,COSA=4/5,且（a-2) : b : (c+2)=1 : 2 : 3判断三角形的形状.
解:
b=2(a-2)
c+2=3(a-2) 即c=3(a-2)-2
cosA=4/5
代入a²=b²+c²-2bccosA中
得a=6,b=8,c=10
c²=a²+b²
△ABC是直角三角形
2.在△ABC中,A=4sin10°,b=2sin50°,C=70° 则S△ABC=?
解:
S=absinC/2
=4sin10sin50sin70
=2sin50[cos(10-70)-cos(10+70)]
=2sin50[cos60-cos80]
=sin50-2sin50cos80
=sin50-[sin(50+80)+sin(50-80)]
=sin50-sin130+sin30
=sin50-sin50+sin30
=1/2