在三角形ABC中，角ABC所对的边分别为abc，已知a=2bsinA，c=根号三b.（1）：求B的 在三角形ABC中角A B C所对的边为abc,已知a=3,b...
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u89d2A\uff0cB\uff0cC\uff0c\u6240\u5bf9\u7684\u8fb9\u4e3aa,b,c.\u5df2\u77e5a=2bsinA.c=\u6839\u53f73b \u82e5\u4e09\u89d2\u5f62\u7684\u9762\u79ef\u4e3a2\u500d\u7684\u6839\u53f73\u6c42ab\u7684\u503c\u6839\u636e\u6b63\u5f26\u5b9a\u7406\u5f97 sinA=2sinBsinA \uff0c
\u6240\u4ee5 sinB=1/2 \uff0c
\u7531\u4e8e c=\u221a3b>b \uff0c\u56e0\u6b64 B \u4e3a\u9510\u89d2\uff0c\u5219 B=\u03c0/6 \uff0c
\u636e\u6b63\u5f26\u5b9a\u7406\uff0csinC=\u221a3sinB=\u221a3/2 \uff0c\u56e0\u6b64 C=\u03c0/3 \uff0c\u6240\u4ee5 A=\u03c0/2 \uff0c
1\uff09\u56e0\u4e3a\u4e09\u89d2\u5f62\u9762\u79efS=1/2*absinC=1/2*a*b*\u221a3/2=2\u221a3 \uff0c
\u89e3\u5f97 a*b=8 \u3002
2\uff09\u56e0\u4e3a\u4e09\u89d2\u5f62\u9762\u79efS=1/2*bc=\u221a3/2*b^2=2\u221a3 \uff0c\u56e0\u6b64\u89e3\u5f97 b=2 \uff0cc=2\u221a3 \uff0c\u6240\u4ee5\u7531\u52fe\u80a1\u5b9a\u7406\u5f97
a=\u221a(b^2+c^2)=4 \u3002
3\uff09AB=c=2\u221a3 \u3002
\uff08\u4e0d\u77e5\u4f60\u8981\u7684\u662f\u54ea\u4e2a\uff0c\u6240\u4ee5\u7ed9\u4e86\u4f60\u591a\u4e2a\u9009\u62e9\uff09
cosA=1/3,\u5219\u6709sinA=2\u6839\u53f72/3
\u6b63\u5f26\u5b9a\u7406\u5f97:a/sinA=b/sinB
sinB=bsinA/a=2*( 2\u6839\u53f72/3)/3=4\u6839\u53f72/9
\u4f59\u5f26\u5b9a\u7406\u5f97:a^2=b^2+c^2-2bccosA
9=4+c^2-4c*1/3
c^2-4c/3-5=0
3c^2-4c-15=0
(3c+5)(c-3)=0
\u6545\u6709c=3
⑴∵a=2bsinA，
由正弦定理得，
sinA=2sinBsinA，
sinB=1/2，
∵c=√3b，
∴c>b，C>B，
∴0<B<π/2，
∴B=π/6；
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⑵c=√3b，
由余弦定理得，
b²=a²+c²-2ac•cosB
∴a²+2b²-3ab=0，①
∵S=1/2•acsinB=√3/2，
∴ab=2，②
联立①②得，
a=2，b=1．
【考点】：三角形正弦定理与余弦定理的综合应用．
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