高中三角形问题 高中三角形问题，很基础但脑子转不过来，求讲解
\u9ad8\u4e2d\u4e09\u89d2\u5f62\u9762\u79ef\u95ee\u9898\uff0c\u5173\u4e8e\u53d6\u503c\u8303\u56f4\uff1fA=\u03c0/3\uff0c\u5219B+C=2\u03c0/3
\u7531\u6b63\u5f26\u5b9a\u7406\u6709\uff1aa/sinA=b/sinB=c/sinC=2/(\u221a3/2)=4/\u221a3
\u6240\u4ee5\uff0cb=(4/\u221a3)sinB\uff0cc=(4/\u221a3)sinC
S\u25b3ABC=(1/2)bcsinA=(1/2)(4/\u221a3)²sinBsinC\u00b7(\u221a3/2)
=(4\u221a3/3)sinBsinC
=(-2\u221a3/3)\u00b7[cos(B+C)-cos(B-C)]
=(-2\u221a3/3)\u00b7[(-1/2)-cos(B-C)]
=(2\u221a3/3)\u00b7[cos(B-C)+(1/2)]
\u56e0\u4e3aB-C\u2208(-2\u03c0/3,2\u03c0/3)\uff0c\u6240\u4ee5cos(B-C)\u2208(-1/2,1]
\u6240\u4ee5\uff0cS\u25b3ABC\u2208(0,\u221a3]
\u7528\u57fa\u7840\u4e0d\u7b49\u5f0f\u601d\u60f3\u6765\u89e3
\u7b2c\u4e00\u4e2a\u540c\u96642 \u5f97 0<B<4\u5206\u4e4b\u5140
\u7b2c\u4e8c\u4e2a\u5c31\u4e0d\u7528\u7ba1
\u7b2c\u4e09\u4e2a\u540c\u65f6\u51cf\u5140\u518d\u4e58 \u8d1f\u4e09\u5206\u4e4b\u4e00 \u4e58\u8d1f\u6570\u8981\u53d8\u53f7 \u5927\u4e8e\u53d8\u5c0f\u4e8e \u5f973\u5206\u4e4b\u5140>B>6\u5206\u4e4b\u5140
\u53d6\u5e76\u96c6
∵180°-C=A+B
由题意:
sinC=(sinA+sinB)/(cosA+cosB)
化简上式：
sin(A+B)=(sinA+sinB)/(cosA+cosB)
sinA+sinB=sin(A+B)(cosA+cosB)
sinA+sinB=(sinAcosB+cosAsinB)(cosA+cosB)
sinA+sinB=sinAcosAcosB+sinA(cosB)^2+(cosA)^2sinB+cosAcosBsinB
sinA[1-(cosB)^2]+sinB[1-(cosA)^2]=(sinA+sinB)cosAcosB
sinA(sinB)^2+sinB(sinA)^2=(sinA+sinB)cosAcosB
sinAsinB(sinA+sinB)=(sinA+sinB)cosAcosB
sinAsinB=cosAcosB
cosAcosB-sinAsinB=0
cos(A+B)=0
cosC=0
C=90º
∴△ABC是直角三角形
(2)
∵(a²-b²)sin(A+B)=(a²+b²)sin(A-B)
根据正弦定理:
[(sinA)^2-(sinB)^2]sin(A+B)=[(sinA)^2+(sinB)^2]sin(A-B)
化简上式:
(sinA)^2*[sin(A+B)-sin(A-B)]=(sinB)^2*[sin(A-B)+sin(A+B)]
(sinA)^2*2cosAsinB=(sinB)^2*2sinAcosB
(sinA)^2*2cosAsinB-(sinB)^2*2sinAcosB=0
sinAsinB(sin2A-sin2B)=0
sin2A=sin2B
①
2A=2B
A=B
△ABC是等腰三角形
②
2A+2B=180°
A+B=90°
C=90°
△ABC是直角三角形
综上所述
△ABC是直角三角形或等腰三角形
绛旓細1銆乥^2+c^2-a^2=鈭3bc cosA=(b^2+c^2-a^2)/2bc=鈭3/2 鎵浠=30搴 2銆乻in(B-C)=sinBcosC-cosBsinC 鎵浠2sinBcosC-sin(B-C)=sinBcosC+cosBsinC =sin(B+C)=sin[180-(B+C)]=sinA =sin30搴 =1/2
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