Question

在三角形ABC中,∠A、∠B、∠C的对边分别为a、b、c,若bcosC=(2a-c) cosB,

(1)求∠B的大小;

(2)若b=,a+c=4,求三角形ABC的面积.

(文)(本小题共12分)在三角形ABC中,三个内角A、B、C的对边分别为a、b、c,且.

(1)求角B的大小;

(2)若b=,a+c=4,求△ABC的面积.

Answer 1

解:(1)由已知及正弦定理可得

sinBcosC=2sinAcosB-cosBsinC,                                              

∴2sinAcosB=sinBcosC+cosBsinC=sin(B+C).

又在△ABC中,sin(B+C)=sinA≠0,                                              

∴2sinAcosB=sinA,

即在△ABC中,cosB=.                                                     

∴B=.                                                                    

(2)∵b²=7=a²+c²-2accosB,

∴7=a²+c²-ac.                                                               

又∵(a+c)²=16=a²+c²+2ac,

∴ac=3.                                                                    

∴S_△ABC=acsinB.

∴S_△ABC=×3×=.                                                  

(文)解:(1)由已知,得

sinBcosC=2sinAcosB-cosBsinC,                                              

∴2sinAcosB=sinBcosC+cosBsinC=sin(B+C).                                    

又在△ABC中,sin(B+C)=sinA≠0,

∴2sinAcosB=sinA,即cosB=.                                              

∵0＜B＜π,∴B=.                                                           

(2)∵b²=7=a²+c²-2accosB=a²+c²-ac,                                           ①

(a+c)²=16=a²+c²+2ac,                                                      ②

由①②可得ac=3,                                                         

∴S_△ABC=acsinB=×3×=.