已知锐角△ABC的内角A,B,C的对边分别为a,b,c,23cos2A+cos2A=0,a=7,c=6,则b=( )
A.10
B.9
C.8
D.5
∵23cos2A+cos2A=23cos2A+2cos2A-1=0,即cos2A=
,A为锐角,1 25
∴cosA=
,1 5
又a=7,c=6,
根据余弦定理得:a2=b2+c2-2bc•cosA,即49=b2+36-
b,12 5
解得:b=5或b=-
(舍去),13 5
则b=5.
故选D
已知锐角△ABC的内角A,B,C的对边分别为a,b,c,23cos2A+cos2A=0,a=7,c=6,则b=( )
A.10
B.9
C.8
D.5
∵23cos2A+cos2A=23cos2A+2cos2A-1=0,即cos2A=
,A为锐角,1 25
∴cosA=
,1 5
又a=7,c=6,
根据余弦定理得:a2=b2+c2-2bc•cosA,即49=b2+36-
b,12 5
解得:b=5或b=-
(舍去),13 5
则b=5.
故选D