在三角形ABC中，角A,B,C所对的边分别为a,b,c,已知a的平方+c的平方=b的平方+ac,且a:c=(√3+1):2，求角C的 三角形ABC中三个内角A，B，C的对边分别为a，b，C，已知...
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u89d2A,B,C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa,b,c\uff0c\u5df2\u77e5(2a\uff0bb)\u00f7c\uff1dcos(A\uff0bC)\u00f7c\u627e\u5230\u539f\u9898\u4e86\uff0c\u4e0b\u9762\u6765\u8865\u5145\u4e00\u4e0b:
\u539f\u9898\u4e3a\uff1a\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u89d2A,B,C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa,b,c\uff0c \u5df2\u77e5(2a\uff0bb)\u00f7c\uff1dcos(A\uff0bC)\u00f7cosC \u6c42C\u7684\u5927\u5c0f \u82e5c \uff1d2\uff0c\u6c42\u4e09\u89d2\u5f62ABC\u9762\u79ef\u6700\u5927\u65f6a,b\u7684\u503c.
(1)\u89e3\uff1a
\u56e0\u4e3a A+B+C=\u03c0\uff1b
\u6240\u4ee5 cos(A+C)=-cosB
\u6240\u4ee5 \u53f3\u5f0f=-cosB/cosC (\u6682\u4e0d\u53ef\u5316\u7b80)
\u6240\u4ee5 \u5de6\u5f0f=(2sinA+sinB)/sinC (\u7528\u6b63\u5f26\u5b9a\u7406\u5316\u7b80)
2sinAcosC+sinBcosC=-cosBsinC ( \u4e24\u8fb9\u540c\u4e58cosCsinC \u53bb\u5206\u6bcd)
\u7531\u4f59\u5f26\u5b9a\u7406 \u5f97\uff1a2sinAcosC=-sin(B+C)
2sinAcosC=-sinA
\u56e0\u4e3a sinA\u22600
\u6240\u4ee5 cosC=-1/2
\u53c8 0<C<\u03c0
\u6240\u4ee5 C=2\u03c0/3
(2)\u89e3\uff1a
\u7531\u4f59\u5f26\u5b9a\u7406\u5f97\uff1a cosC=-1/2=\uff08a^2+b^2-c^2\uff09/(2ab)
4-ab=a^2+b^2\u300b2ab \uff08\u57fa\u672c\u4e0d\u7b49\u5f0f\uff09
ab\u300a4/3
\u5f53\u4e14\u4ec5\u5f53 a=b=\u4e09\u5206\u4e4b\u4e8c\u6839\u53f7\u4e09 \u65f6 \u53d6\u7b49
\u53c8 S=1/2 absinC
S\u6700\u5927\u65f6\uff0cab\u53d6\u6700\u5927\u503c4/3
\u6240\u4ee5 S\u6700\u5927\u65f6\uff0ca=b=\u4e09\u5206\u4e4b\u4e8c\u6839\u53f7\u4e09
\u7531a²+c²=b²+ac\u53d8\u5f62\uff0c\u5f97(a²+c²-b²)/2ac=1/2\u7531\u4f59\u5f26\u5b9a\u7406\uff0c\u5f97\uff1acosB=1/2B=60\u00b0\u6545A+C=120\u00b0\u7531\u6b63\u5f26\u5b9a\u7406a/c=sinA/sinC=sin(120\u00b0-C)/sinC==(\u221a3+1)/2sin(120\u00b0-C)/sinC=(\u221a3cosC/2+sinC/2)/sinC=\u221a3cotC/2+1/2=(\u221a3+1)/2\u89e3\u5f97\uff1acotC=1C=45\u00b0
由a^2+c^2=b^2+ac得a^2+c^2-b^2=ac,
由余弦定理得cosB=(a^2+c^2-b^2)/(2ac)=ac/(2ac)=1/2
故有B=60,A+C=180-B=120.A=120-C.
再由正弦定理得sinA/sinC=a/c=(√3+1)/2
2sinA=(√3+1)sinC,2sin(120-C)=(√3+1)sinC
2sin120cosC-2sinCcos120=(√3+1)sinC
√3cosC+sinC=(√3+1)sinC
√3cosC=√3sinC
tanC=1,故得C=45
绛旓細绠鍗曞垎鏋愪竴涓嬶紝绛旀濡傚浘鎵绀
绛旓細鈭礏涓涓夎褰鐨勫唴瑙掞紝涓攃osB=4/5>0锛屸埓B涓洪攼瑙掞紝鈭磗inB=鈭氾蓟1-(cosB)^2]=3/5,鈶存牴鎹寮﹀畾鐞嗗緱锛歛/sinA=b/sinB锛屸埓 sinA=2脳3/5梅3=2/5锛屸懙鈭礱<b,锛孊涓洪攼瑙掞紝鈭碅涓洪攼瑙掞紝S=1/2ac sinB=3锛屸埓c=5锛宐=鈭歔c^2+a^2-2ac cosB锛=鈭(25+4-16)=鈭13銆....
绛旓細瑕佷娇涓夎褰㈡湁涓よВ,灏辨槸瑕佷娇浠涓哄渾蹇,鍗婂緞涓2鐨勫渾涓嶣A鏈変袱涓氦鐐,褰瑙扐绛変簬90鏃剁浉鍒,褰撹A绛変簬45鏃朵氦浜嶣鐐,涔熷氨鏄彧鏈変竴瑙.鎵浠ヨA澶т簬45灏忎簬90.鏍瑰彿2/2<sinA<1 鐢辨寮﹀畾鐞嗭細a*sinB=b*sinA.浠ｅ叆寰楀埌锛歛=x=b*sinA/sinB=2鍊嶇殑鏍瑰彿2*sinA 2<a<2鍊嶇殑鏍瑰彿2 ...
绛旓細A+B+C=3B=180 B=60 cosB=(a^2+c^2-b^2)/2ac=1/2 a^2+c^2-ac=ac (a-c)^2=0 a=c A=C A+B+C=180 A=C=60 鈭涓夎褰BC涓虹瓑杈逛笁瑙掑舰
绛旓細鍒╃敤姝ｅ鸡瀹氱悊.a/sinA=b/sinB=c/sinC c/b=sinC/sinB=sin2B/sinB=2sinBcosB/sinB=2cosB
绛旓細(1) 鈭C鏄挐瑙 鈭碅銆B鏄攼瑙 sinA=3/5 cosA=鈭(1-sin^2A)=4/5 tan(A-B)=16/63 (tanA-tanB)/(1+tanAtanB)=16/63 (sinA/cosA-sinB/cosB)/[1+sinAsinB/(cosAcosB)]=16/63 [(sinAcosB-cosAsinB)/(cosAcosB)]/[(cosAcosB+sinAsinB)/(cosAcosB)]=16/63 (sinAcosB-cosAsinB)/...
绛旓細4cos²B+4cosB-3=0 (2cosB-1)(2cosB+3)=0 鈭碿osB=1/2锛屾垨cosB=-3/2 (鑸嶅幓)鈭碆=蟺/3 (2) cosB=1/2锛宻inB=鈭3/2锛宑osA=鈭13/13锛宻inA=2鈭39/13 鈭磗inC=sin(A+B)=sinAcosB+cosAsinB=3鈭39/26 姝ｅ鸡瀹氱悊锛歝/sinC=a/sinA锛屸埓a=4 鈭碨鈻ABC=1/2*ac*sinB=3鈭3 ...
绛旓細绛旓細1锛涓夎褰BC涓锛bcosA+acosB=-2ccosC 鏍规嵁姝ｅ鸡瀹氱悊鏈夛細a/sinA=b/sinB=c/sinC=2R 鎵浠ワ細sinBcosA+sinAcosB=-2sinCcosC 鎵浠ワ細sin(A+B)=-2sinCcosC=sinC>0 鎵浠ワ細cosC=-1/2 瑙ｅ緱锛欳=120掳 2锛変笁瑙掑舰ABC闈㈢НS=(ab/2)sinC=2鈭3 鎵浠ワ細absin120掳=4鈭3 瑙ｅ緱锛歛b=8 ...
绛旓細绛旓細涓夎褰BC婊¤冻c=2acosB 鏍规嵁浣欏鸡瀹氱悊鏈夛細cosB=(a^2+c^2-b^2)/(2ac)=c/(2a)a^2+c^2-b^2=c^2 a^2=b^2 a=b 鎵浠ワ細涓夎褰㈡槸绛夎叞涓夎褰
绛旓細鐢变綑寮﹀畾鐞嗗緱cosB=(a^2+c^2-b^2)/(2ac)=ac/(2ac)=1/2 鏁呮湁B=60,A+C=180-B=120.A=120-C.鍐嶇敱姝ｅ鸡瀹氱悊寰梥inA/sinC=a/c=(鈭3+1)/2 2sinA=(鈭3+1)sinC,2sin(120-C)=(鈭3+1)sinC 2sin120cosC-2sinCcos120=(鈭3+1)sinC 鈭3cosC+sinC=(鈭3+1)sinC 鈭3cosC=鈭3...
