在△ABC中,内角A,B,C所对的边长分别是a,b,c 在△ABC中，内角A,B,C所对的边分别为a,b,c,已知3...
\u5728\u25b3ABC\u4e2d\uff0c\u89d2A\uff0cB\uff0cC\u6240\u5bf9\u7684\u8fb9\u957f\u5206\u522b\u662fa\uff0cb\uff0cc,2cosC(acosC+ccosA)+b=01\u3001\u6839\u636e\u6b63\u5f26\u5b9a\u7406\uff0ca/sinA=b/sinB=c/sinC=k
\u6240\u4ee5a=ksinA\uff0cb=ksinB\uff0cc=ksinC
2cosC(ksinAcosC+kcosAsinC)+ksinB=0
2cosCsin(A+C)+sinB=0
2cosCsinB+sinB=0
sinB(2cosC+1)=0
cosC=-1/2
C=2\u03c0/3
2\u3001\u6839\u636e\u4f59\u5f26\u5b9a\u7406\uff0cc^2=a^2+b^2-2abcosC
12=a^2+4-2a*2*(-1/2)=a^2+2a+4
a^2+2a-8=0
(a+4)(a-2)=0
a=2\u6216-4\uff08\u820d\u53bb\uff09
\u6240\u4ee5S\u25b3ABC=(1/2)*absinC
=(1/2)*2*2*\u221a3/2
=\u221a3
\u89e3\uff1a\u7531\u9898\u610f\uff0c\u53ef\u77e5
A\u4e3a\u9510\u89d2
\u2235sinA=\u221a10/10
\u2234cosA=\u221a(1-sin²A)=3\u221a10/10
\u2235sinC=sin[\u03c0-(A+B)]=sin(A+B)
\u2234sinC=sinAcosB+cosAsinB
=(\u221a10/10)\u00d7cos(\u03c0/4)+(3\u221a10/10)\u00d7sin(\u03c0/4)
=2\u221a5/5
\u2235a/sinA=b/sinB
\u2234a:b=sinA:sinB=(\u221a10/10)\u00f7sin(\u03c0/4)=\u221a5/5
\u540c\u7406\uff0c\u53ef\u5f97b:c=\u221a10/4
\u2234a:b:c=\u221a2:\u221a10:4
\u4ee4a=\u221a2k (k>0)
\u5219b=\u221a10k
\u2234S=(1/2)absinC
\u22349=(1/2)\u00d7\u221a2k\u00d7\u221a10k\u00d7(2\u221a5/5)
\u22342k²=9
\u6545k=3\u221a2/2
\u2234a=\u221a2k=\u221a2\u00d7(3\u221a2/2)=3
2√3sin(A+B)=4sinAsinB
2√3sin(π/3)=4sinAsinB
3=4sinAsinB
∠B=180°-∠C-∠A
sinB=sin(C+A)
=sinCcosA+cosCsinA
=√3cosA/2+sinA/2
3=4sinAsinB
=4sinA(√3cosA/2+sinA/2)
=2√3sinAcosA+2sinAsinA
2√3sinAcosA+(sinA)^2-(cosA)^2=2
√3sin2A-cos2A=2
sin(2A-π/6)=1
2A-π/6=π/2
A=π/3
B=π/3
∠A=∠B=∠C,a=b=c=2
(2)前面未用这个条件的时候,已经解出,是等边三角形
不过,用这个条件可以验算一下:
sin C + sin(B - A)= sin2A
sinC=sin(A+B)
sinAcosB+cosAsinB+sinBcosA-cosBsinA=2sinAcosA
cosAsinB=sinAcosA
sinB=sinA
结果是一样的.
若a∧2-b∧2=√3 bc，sinC=2√3 sinB，求A
解：
由余弦定理
cosA=(b²+c²-a²)/2bc
=[c²-(a²-b²]/2bc
=[c²-(√3)bc]/2bc
=c/(2b)-(1/2)√3 (*)
由正弦定理
c/b=sinC/sinB=2√3
代入(*)得
cosA=(√3)/2
∵ 0<A<180º
∴ A=30º
数据有问题，我算出来根号下还有根号
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绛旓細瑙ｏ細(1)鐢a/sinA=b/sinB锛寰梐sinB=bsinA銆傚張asinA=4bsinB锛屽緱4bsinB=asinA銆備袱寮忎綔姣斿緱锛歛/4b=b/a 鈭碼=2b锛庣敱ac=鏍瑰彿5(a²-b²-c²)锛屽緱b²+c²-a²=-鈭5/5ac 鐢变綑寮﹀畾鐞嗭紝寰 cosA=b²+c²-a²/2bc=-鈭5/5ac/ac=-5/...
绛旓細A=120掳锛宐=1,涓斾笁瑙掑舰ABC闈㈢Н涓烘牴鍙3锛孲螖=1/2*bcsinA=鈭3 鈭1/2*c*鈭3/2=鈭3 鈭碿=4 鏍规嵁浣欏鸡瀹氱悊 a�0�5=b�0�5+c�0�5-2bccosA =1+16+4=21 鈭碼=鈭21 鏍规嵁姝ｅ鸡瀹氱悊锛2R=a/sinA=鈭21/(鈭3/2)=2鈭7 鈭(a+...
绛旓細瑙ｏ細A銆丅銆丆涓鈻矨BC鐨勪笁涓鍐呰锛屽垯A+B+C=蟺锛孉=蟺-(B+C)cosC=1/3,鍒檚inC=鈭(1-cos²C)=2鈭2/3 a,b,c涓衡柍ABC涓鍐呰鎵瀵圭殑杈癸紝a=2b锛屾牴鎹寮﹀畾鐞嗘湁a/sinA=b/sinB锛屽垯2b/sinA=b/sinB,鍗硈inA=2sinB sinA=sin[蟺-(B+C)]=sin(B+C)=2sinB ,鍗硈inBcosC+cosBsinC=...
绛旓細鍦ㄢ柍ABC涓,鍐呰A,B,C鎵瀵圭殑杈逛緷娆′负a,b,c,濡傛灉婊¤冻B=30掳,b=4鐨勨柍ABC鎭版湁涓涓紝鍒檃鐨勫彇鍊艰寖鍥存槸锛氳В锛氬洓涓瓟妗堥兘涓嶅锛佹纭瓟妗堟槸4<a鈮4鈭3.杩欎釜闂鏈濂界敤浣滃浘娉曟眰瑙ｃ傛敞鎰忔潯浠讹細B=30掳锛宐=4鐨勨柍ABC鎭版湁涓涓傜壒鍒敞鎰忊滄伆鏈変竴涓濄備綔鈭燲BY=30掳锛涢《鐐笰鍦ㄥ皠绾緽Y涓婄Щ鍔紝浠 ...
绛旓細鐢变綑寮﹀畾鐞 cosA=(b²+c²-a²)/2bc =[c²-(a²-b²]/2bc =[c²-(鈭3)bc]/2bc =c/(2b)-(1/2)鈭3 (*)鐢辨寮﹀畾鐞 c/b=sinC/sinB=2鈭3 浠ｅ叆(*)寰 cosA=(鈭3)/2 鈭 0<A<180º鈭 A=30º...
绛旓細瑙ｏ細锛1锛夆埖鍦ㄢ柍ABC涓紝鏍规嵁姝ｅ鸡瀹氱悊寰梐sinA=bsinB锛鈭碽sinA=asinB锛庡張鈭电敱宸茬煡寰梑sinA=-鈭3acosB锛屸埓sinB=-鈭3cosB锛屽彲寰梩anB=-鈭3锛屸埖鍦ㄢ柍ABC涓紝0锛淏锛溝锛屸埓B=2蟺3锛涳紙2锛夛紙鈪帮級鈭礏D涓衡垹ABC鐨勫钩鍒嗙嚎锛屸埓鈭燗BD=鈭燙BD=蟺3锛庘埖S鈻矨BC=S鈻矪CD+S鈻矨BD锛孊D=1銆丅C=x涓擝A=y锛...
绛旓細鈭a锛宐锛宑鎴愮瓑姣旀暟鍒楋紝鈭碽2=ac锛庣敱浣欏鸡瀹氱悊鍙緱锛歜2=a2+c2-2ac?cos60掳锛屸埓ac=a2+c2-ac锛岃В寰梐=c锛庡張鈭燘=60掳锛屸埓鈻矨BC涓虹瓑杈逛笁瑙掑舰锛庢晠閫夛細C锛
绛旓細鈶′綑寮﹀畾鐞嗭細a²=b²+c²-2bccosA=b²+c²-鈭2*bc=16锛屸埓b²+c²=鈭2*bc+16 鑰宐²+c²鈮2bc锛屸埓鈭2*bc+16鈮2bc锛屸埓bc鈮16/(2-鈭2)=8(2+鈭2)鈭碨鈻矨BC=1/2*bc*sinA=鈭2/4*bc鈮も垰2/4*8(2+鈭2)=4+4鈭2 ...
绛旓細瑙ｏ細鐢变綑寮﹀畾鐞嗗緱 cosA=b^2+c^2-a^2/2bc cosC=a^2+b^2-c^2/2ab 鍥犱负A=蟺/4=45搴 b^2-a^2=c/2^2 鎵浠ユ牴鍙2/2=3c^2/2/2bc c=2鍊嶆牴鍙2b/3 c^2=8b^2/9 a^2=5b^2/9 a=鏍瑰彿5b/3 鎵浠osC=鏍瑰彿5/5 鎵浠osC鐨勫兼槸鏍瑰彿5/5 ...
