设△ABC的内角A,B,C,所对的边分别为a,b,c,且b=3,c=1,S△ABC=根号2求cosA与a的值

Question

设△ABC的内角A,B,C,所对的边分别为a,b,c,且b=3,c=1,S△ABC=根号2
求cosA与a的值

Answer 1

解由SΔABC=1/2bcsinA
=1/2*3*1*sinA=√2
即sinA=2√2/3
故cosA=±√(1-sin^2A)=±1/3
当cosA=1/3时,
a^2=b^2+c^2-2bccosA
=3^2+1^2-2*3*1*1/3
=10-2
=8
即a=2√2
当cosA=-1/3时,即A是钝角
a^2=b^2+c^2-2bccosA
=3^2+1^2-2*3*1*(-1/3)
=10+2
=12
即a=2√3
故综上知cosA=1/3,a=2√2或cosA=-1/3,a=2√3.