△ABC中已知边a，b与∠A怎么求∠C
\u5df2\u77e5\u25b3ABC\u4e2d\uff0c\u2220A\uff0c\u2220B\uff0c\u2220C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa\uff0cb\uff0cc\uff0c\u4e0b\u5217\u6761\u4ef6\u4e2d\u80fd\u5224\u65ad\u51fa\u662f\u25b3ABC\u76f4\u89d2\u4e09\u89d2\u5f62\u7684\u6709\uff08\u3000\u3000\uff09\uff08\uff081\uff09\u2235\u2220A\uff1a\u2220B\uff1a\u2220C=3\uff1a4\uff1a5\uff0c\u2234\u8bbe\u2220A=3x\uff0c\u5219\u2220B=4x\uff0c\u2220C=5x\uff0c\u22343x+4x+5x=180\u00b0\uff0c\u89e3\u5f97x=15\u00b0\uff0c5x=15\u00d75=75\u00b0\uff0c\u2234\u6b64\u4e09\u89d2\u5f62\u662f\u9510\u89d2\u4e09\u89d2\u5f62\uff0c\u6545\u672c\u5c0f\u9898\u9519\u8bef\uff1b\uff082\uff09\u223552+122=132\uff0c\u2234\u662f\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u6545\u672c\u5c0f\u9898\u6b63\u786e\uff1b\uff083\uff09\u2235102+112\u2260122\uff0c\u2234\u6b64\u4e09\u89d2\u5f62\u4e0d\u662f\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u6545\u672c\u5c0f\u9898\u9519\u8bef\uff1b\uff084\uff09\u2235a2+b2+c2+50=6a+8b+10c\uff0c\u2234a2-6a+9+b2-8b+16+c2-10c+25=0\uff0c\u5373\uff08a-3\uff092+\uff08b-4\uff092+\uff08c-5\uff092=0\uff0c\u2234a=3\uff0cb=4\uff0cc=5\uff0c\u223532+42=52\uff0c\u2234\u25b3ABC\u662f\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u6545\u672c\u5c0f\u9898\u6b63\u786e\uff1b\uff085\uff09\u2235\u539f\u5f0f\u53ef\u5316\u4e3a\uff1aa2+c2+2ac=b2+2ac\uff0c\u5373a2+c2=b2\uff0c\u2234\u25b3ABC\u662f\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u6545\u672c\u5c0f\u9898\u6b63\u786e\uff1b\uff086\uff09\u2235\u2220A-\u2220B=\u2220C\uff0c\u2234\u2220A-\u2220B-\u2220C=0\u2460\uff0c\u2235\u2220A+\u2220B+\u2220C=180\u00b0\u2461\uff0c\u2234\u2460+\u2461\u5f97\uff0c2\u2220A=180\u00b0\uff0c\u89e3\u5f97\u2220A=90\u00b0\uff0c\u2234\u25b3ABC\u662f\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u6545\u672c\u5c0f\u9898\u6b63\u786e\uff0e\u6545\u9009C\uff0e
:\u56e0\u4e3aa/b=( a+b)/(a+b+c),
\u6240\u4ee5a(a+b+c)=b(a+b)\u5373b^2=a^2+ac
\u5bf9\u4e8e\u4e09\u89d2\u5f62ABC,\u5ef6\u957fCB\u81f3\u70b9D\uff0c\u4f7f\u5f97BD=BA=c
\u56e0\u4e3ab^2=a^2+ac
\u53c8\u56e0\u4e3a\u89d2BCA=\u89d2ACD,
\u6240\u4ee5\u4e09\u89d2\u5f62ACB\u4e0e\u4e09\u89d2\u5f62DCA\u76f8\u4f3c
\u6240\u4ee5\u89d2CDA\uff1d\u89d2DAB\uff1d\u89d2CAB
\u6240\u4ee5\u89d2CBA=2\u89d2CAB
\u53e6\u4e00\u79cd\u89e3\u6cd5\uff1a\u9ad8\u4e2d\u89e3\u6cd5\u5982\u4e0b:
a/b=( a+b)/(a+b+c)\u5316\u7b80\u53ef\u5f97b2 \u2013 a2 = ac
\u4f59\u5f26\u516c\u5f0f\uff1acosA = (b2+c2 - a2 ) / 2bc
cosB = (a2+c2 \u2013b2 ) / 2ac
\u5e26\u5165\u9898\u76ee\u6761\u4ef6\u5316\u7b80\u5f97 cosA =(c + a ) / 2b
cosB =(c \u2013 a ) / 2a
\u53ef\u77e5 c = 2b*cosA \u2013 a = 2a * cosB + a
\u5316\u7b80\u4e00\u4e0b\u5f97\uff1ab *cosA \u2013 a*cosB = a
\u7528\u6b63\u5f26\u5b9a\u7406\u5e26\u5165
sinB*cosA \u2013 sinA*conB = sinA
\u6240\u4ee5\u6709 sin(B \u2013A ) = sinA
\u6240\u4ee5\u5f97B-A=A \u6216 B \u2013 A = 180-A\uff08\u540e\u8005B = 180\uff0c\u4e0d\u53ef\u80fd\uff09
\u6240\u4ee5 B= 2A
采纳哟
先用余弦定理求出c边。
然后用余弦定理求出∠C
如有疑问，请追问；如已解决，请采纳
用正弦定理求出∠b。
利用三角形内角和180°求出∠c
根据正弦定理，求出B，
sinB/b＝sinA/a,B＝arcsin(bsinA/a)
C＝π-B-A
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