三角形abc中，内角A.B.C的对边分别为a.b.b。且√3bsinA=acosB （1）求角B大小 （2）若b=√3,a=3,求三...
\u5728\u4e09\u89d2\u5f62ABC\u4e2d,\u5185\u89d2A,B,C\u7684\u5bf9\u8fb9\u5206\u522b\u4e3aa,b,c,\u4e14bsinA=\u6839\u53f73acosB,\u6c42\u89d2B\u7684\u5927\u5c0f,\u82e5b=3,sinC=2sinA,\u6c42a,c\u7684\u503cbsinA=\u221a3acosB
a/sinA=\u221a3b/3cosB
\u56e0\u4e3a a/sinA=b/sinB
\u6240\u4ee5\u221a3b/3cosB=b/sinB
\u221a3sinB=3cosB
1/2sinB-\u221a3/2cosB=0
sin(B-\u03c0/3)=0
B=\u03c0/3
(2)
sinC=2sinA,\u5373\u6709c=2a
b^2=a^2+c^2-2accosB
9=a^2+4a^2-2a*2a*1/2
9=5a^2-2a^2
a^2=3
a=\u6839\u53f73
c=2a=2\u6839\u53f73
\uff08\u2160\uff09\u5728\u25b3ABC\u4e2d\uff0c\u2235bsinA=3acosB\uff0c\u7531\u6b63\u5f26\u5b9a\u7406\u53ef\u5f97 sinBsinA=3sinAcosB\uff0c\u6545\u6709tanB=3\uff0c\u2234B=\u03c03\uff0e\uff08\u2161\uff09\u82e5a=2\uff0cc=1\uff0eD\u4e3aAC\u4e2d\u70b9\uff0c\u7531\u4f59\u5f26\u5b9a\u7406\u53ef\u5f97 b2=a2+c2-2ac?cosB=4+1-2\u00d72\u00d71\u00d712=3\uff0c\u2234b2+c2=a2\uff0c\u2234A=\u03c02\uff0e\u2235AD=12b=32\uff0c\u2234BD=AB2+AD 2=72\uff0e
解由√3bsinA=acosB由正弦定理知
√3sinBsinA=sinAcosB
即
√3sinB=cosB
即
√3=cosB/sinB=cotB
cotB=√3
即B=30°
2 由余弦定理知
b²=a²+c²-2accosB
即
(√3)²=3²+c²-2*3ccos30°
即c²-3√3c+6=0
解得c=2√3或c=√3
当c=2√3时，SΔABC=1/2acsinB=1/2*3*2√3*sin30°=3√3/2
当c=√3时，SΔABC=1/2acsinB=1/2*3*√3*sin30°=3√3/4.
(1)√3bsinA=acosB，b/a=cosB/√3sinA,
由正弦定理：b/a=sinB/sinA= cosB/√3sinA,
tanB=1/√3,
B=30°
(2)由上面知：c=b,C=B=30°,
三角形面积=1/2*absinC=1/2*3*√3*sin30°=3/4*√3
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