三角形ABC的内角A,B,C的对边分别为a,b,c 三角形ABC的内角A,B,C的对边分别为a,b,c.已知si...
\u4e09\u89d2\u5f62ABC\u7684\u5185\u89d2A,B,C\u7684\u5bf9\u8fb9\u5206\u522b\u4e3aa,b,c,\u5df2\u77e5a=bcosC+csinB\u89e3\u7b54\uff1a
\uff081\uff09
\u5229\u7528\u6b63\u5f26\u5b9a\u7406:a/sinA=b/sinB=c/sinC
\u2235 a=bcosC+csinB
\u2234 sinA=sinBcosC+sinCsinB
\u2235 sinA=sin[\u03c0-(B+C)]=sin(B+C)
\u2234 sinBcosC+cosCsinB=sinBcosC+sinCsinB
\u2234 cosCsinB=sinCsinB
\u2234 tanB=1
\u2234 B=\u03c0/4
\uff082\uff09
S=(1/2)acsinB=(\u221a2/4)ac
\u5229\u7528\u4f59\u5f26\u5b9a\u7406
4=a²+c²-2ac*cos(\u03c0/4)
\u2234 4=a²+c²-\u221a2ac\u22652ac-\u221a2ac
\u2234 ac\u22644/(2+\u221a2)=2(2+\u221a2)
\u5f53\u4e14\u4ec5\u5f53a=c\u65f6\u7b49\u53f7\u6210\u7acb
\u2234 S\u7684\u6700\u5927\u503c\u662f(\u221a2/4)*2*(2+\u221a2)=\u221a2+1
\u7b80\u4ecb
\u4e09\u89d2\u5f62(triangle)\u662f\u7531\u540c\u4e00\u5e73\u9762\u5185\u4e0d\u5728\u540c\u4e00\u76f4\u7ebf\u4e0a\u7684\u4e09\u6761\u7ebf\u6bb5\u2018\u9996\u5c3e\u2019\u987a\u6b21\u8fde\u63a5\u6240\u7ec4\u6210\u7684\u5c01\u95ed\u56fe\u5f62\uff0c\u5728\u6570\u5b66\u3001\u5efa\u7b51\u5b66\u6709\u5e94\u7528\u3002
\u5e38\u89c1\u7684\u4e09\u89d2\u5f62\u6309\u8fb9\u5206\u6709\u666e\u901a\u4e09\u89d2\u5f62\uff08\u4e09\u6761\u8fb9\u90fd\u4e0d\u76f8\u7b49\uff09\uff0c\u7b49\u8170\u4e09\u89d2\uff08\u8170\u4e0e\u5e95\u4e0d\u7b49\u7684\u7b49\u8170\u4e09\u89d2\u5f62\u3001\u8170\u4e0e\u5e95\u76f8\u7b49\u7684\u7b49\u8170\u4e09\u89d2\u5f62\u5373\u7b49\u8fb9\u4e09\u89d2\u5f62\uff09\uff1b\u6309\u89d2\u5206\u6709\u76f4\u89d2\u4e09\u89d2\u5f62\u3001\u9510\u89d2\u4e09\u89d2\u5f62\u3001\u949d\u89d2\u4e09\u89d2\u5f62\u7b49\uff0c\u5176\u4e2d\u9510\u89d2\u4e09\u89d2\u5f62\u548c\u949d\u89d2\u4e09\u89d2\u5f62\u7edf\u79f0\u659c\u4e09\u89d2\u5f62\u3002
\u25b3ABC\uff0csinB=sin(A+C)=sinAcosC+cosAsinC
\u6240\u4ee5\u6709sinAsinC+cosAsinC=0=(sinA+cosA)sinC
=\u221a2sin(A+\u220f/4)sinC\uff0c\u25b3ABC\uff0c\u220f>C>0\uff0c\u220f>A>0\u6240\u4ee5A+\u220f/4=\u220f\uff0cA=3\u220f/4\u3002
\u6240\u4ee5sinC/c=sinA/a=sinC/\u221a2=(\u221a2/2)/2\uff0c
sinC=1/2\uff0c\u25b3ABC\uff0cA=3\u220f/4\uff0c\u6240\u4ee5C=\u220f/6\u3002
\u540c\u89d2\u4e09\u89d2\u51fd\u6570
\uff081\uff09\u5e73\u65b9\u5173\u7cfb\uff1a
sin^2(\u03b1)+cos^2(\u03b1)=1
tan^2(\u03b1)+1=sec^2(\u03b1)
cot^2(\u03b1)+1=csc^2(\u03b1)
\uff082\uff09\u79ef\u7684\u5173\u7cfb\uff1a
sin\u03b1=tan\u03b1*cos\u03b1 cos\u03b1=cot\u03b1*sin\u03b1
tan\u03b1=sin\u03b1*sec\u03b1 cot\u03b1=cos\u03b1*csc\u03b1
sec\u03b1=tan\u03b1*csc\u03b1 csc\u03b1=sec\u03b1*cot\u03b1
ac=b^2-a^2，b^2=a(a+c)
根据余弦定理有：b^2=a^2+c^2-2accosB
即：a^2+ac=a^2+c^2-2accosB
cosB=(c-a)/(2a)=c/a/2-1/2……（1）
根据正弦定理有：a/sinA=b/sinB
a^2/b^2=(sinA/sinB)^2
a^2/[a(a+c)]={[sin(π/6)]^2}/(sinB)^2
(sinB)^2=(a+c)/(4a)=1/4+c/a/4……(2)
由(1)和（2）知：
c/a=2cosB+1=4(sinB)^2-1=4-4(cosB)^2-1
2(cosB)^2+cosB-1=0
cosB=1/2或者cosB=-1
因为0°<B<180°，所以cosB=-1不符合。
故：cosB=1/2
所以：B=π/3
解：由余弦定理，得：
2bccosA=b^2+c^2-a^2.
∵ ac=b^2-a^2.
∴ 2bccosA=ac+c^2. c≠0.
2bcosA=a+c.
2sinBcosA=sinA+sinC. [由正弦定理得：a=2RsinA, b=2RsinB,..]
2sinBcosA=sinA+sin(A+B). [sinC=sin[π-(A+B)=sin(A+B)]
2sinBcosA=sinA+sinAcosB+cosAsinB.
sinBcosA-cosBsinA=sinA.
sin(B-A)=sinA.
∴B-A=A (1)
或，B-A=π-A (2).
由(2)得：B=π ,显然不合题设要求，故舍去。
由(1)得：B=2A=2*π/6.
∴∠B=π/3.
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绛旓細璁鹃攼瑙涓夎褰BC鐨勫唴瑙扐,B,C鐨瀵硅竟鍒嗗埆涓篴,b,c,a^2+c^2-b^2=鏍瑰彿3ac 姹俢osA+sinC鐨勫彇鍊艰寖鍥 a²+c²-b²=鈭3ac 鐢变綑寮﹀畾鐞嗗緱 cosC=(a²+c²-b²)/(2ac)=鈭3/2 C=蟺/6 A+B+C=蟺锛孉=蟺-C-B=蟺-蟺/6-B=5蟺/6-B A>0锛孊>0锛 0...
绛旓細鈭磗in(C+蟺/3)=0 C+蟺/3=蟺 C=2蟺/3 (2) 鈭礐=2蟺/3 鈭碅+B=蟺/3 sinA=鈭(1-cos²A)=鈭(1-(3/5)²)=4/5 sin(A+B)=sin蟺/3 sinAcosB+cosAsinB=鈭3/2 4/5鈭(1-sin²B)+3/5sinB=鈭3/2 8鈭(1-sin²B)+6sinB=5鈭3 8鈭(1-sin²...
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绛旓細鏍规嵁宸茬煡 a=3 b=4 sinA=3/5 ,鍙煡c=5 鏁卌osA=4/5 cos2A=cosA^2-sinA^2=7/25 (鍏跺疄濡傛灉浣犲仛棰樺鐨勮瘽鐚滈兘鑳界寽鍑洪偅鏄竴涓洿瑙涓夎褰)
绛旓細鈥斺斻(2c-a)/b=(4RsinC-2RsinA)/2RsinB=(2sinC-sinA)/sinB=(cosA-2cosC)/cosB锛鈥斺斻媍osB(2sinC-sinA)=sinB(cosA-2cosC)锛屸斺斻2(cosBsinC+sinBcosC)=cosAsinB+sinAcosB锛屸斺斻2sin(B+C)=2sinA=sin(A+B)=sinC锛屸斺斻媠inC/sinA=2 锛2锛夈乧osC=鈭(1-sin^2C)=鈭...
绛旓細棰樼洰鈥渂2=ac鈥,搴旇鏄痓�0�5=ac鍚э紵 瑙ｏ細鐢ㄤ綑寮﹀畾鐞嗭細cosB=(a�0�5+c�0�5-b�0�5)/2ac 鎶奵=2a甯﹀叆寰梒osB=锛坅�0�5+4a�0�5-ac锛/2ac 鏁寸悊锛歝osB=锛5a�0&...
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