解cosA=(b²+c²-a²)/2bc
=(2²+(√6+√2)²-(2√2)²)/2*2*(√6+√2)
=(4+8+4√3-8)/2*2*(√6+√2)
=(√3+1)/√2(√3+1)
=√2/2
即A=45°
由a/sinA=b/sinB
即2√2/sin45°=2/sinB
sinB=1/2
即B=30°或B=150°
由A=45°,即B<135°,故B=150°舍去
即B=30°
即C=180°-A-B=180°-45°-30°=105°
即A=45°,B=30°,C=105°
解cosA=(b²+c²-a²)/2bc
=(2²+(√6+√2)²-(2√2)²)/2*2*(√6+√2)
=(4+8+4√3-8)/2*2*(√6+√2)
=(√3+1)/√2(√3+1)
=√2/2
即A=45°
由a/sinA=b/sinB
即2√2/sin45°=2/sinB
sinB=1/2
即B=30°或B=150°
由A=45°,即B<135°,故B=150°舍去
即B=30°
即C=180°-A-B=180°-45°-30°=105°
即A=45°,B=30°,C=105°