在锐角三角形ABC中,已知角A=2角C,则a/c的取值范围是
\u5728\u9510\u89d2\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u5df2\u77e5A=2C\uff0c\u5219a/c\u7684\u8303\u56f4\u662f\uff1f\u89e3\uff1a(1) \u7531A=2C,A+B+C=180º,\u53ef\u5f97B=180º-3C.\u53c8\u22bfABC\u4e3a\u9510\u89d2\u4e09\u89d2\u5f62\uff0c\u65450\uff1cB\uff1c90º.\u5373\u67090\uff1c180º-3C\uff1c90º.===>30º\uff1cC\uff1c60º.===>1/2\uff1ccosC\uff1c\u221a3/2.(2)\u7531A=2C.===>sinA=sin2C=2sinCcosC,===>cosC=sinA/(2sinC).\u518d\u7531\u4f59\u5f26\u5b9a\u7406\u53ef\u77e5\uff0ca/c=sinA/sinC.\u6545cosC=a/(2c).\u7ed3\u5408\u524d\u9762\u53ef\u77e5\uff1a1/2\uff1ca/(2c)\uff1c\u221a3/2.===>1\uff1ca/c\uff1c\u221a3.\u5373a/c\u2208(1,\u221a3).
\u89e3\u7b54\uff1a
\u5148\u786e\u5b9a\u2220B\u7684\u8303\u56f4
\u2220A=2\u2220B<\u03c0/2 , \u2234 \u2220B<\u03c0/4
\u2220C=\u03c0-\u2220A-\u2220B=\u03c0-3\u2220B\u03c0/6
\u5373 \u03c0/6 < \u2220B < \u03c0/4
\u5229\u7528\u6b63\u5f26\u5b9a\u7406a/sinA=b/sinB=c/sinC
\u2234 c/b
=sinC/sinB
=sin(\u03c0-3B)/sinB
=sin(3B)/sinB
=sin(2B+B)/sinB
=(sin2BcosB+cos2BsinB)/sinB
=2cos²B+cos2B
=2cos²B+2cos²B-1
=4cos²B-1
\u2235 \u03c0/6 < \u2220B < \u03c0/4
\u2234 cosB\u2208(\u221a2/2,\u221a3/2)
cos²B\u2208(1/2,3/4)
\u2234 4cos²B-1\u2208(1,2)
绛旓細sinA=2鈭2/3 閿愯鈻ABC 鎵浠osA=鈭歔1-(sinA)^2]=1/3 cosA=1-2[sin(A/2)]^2 鎵浠sin(A/2)]^2=1/3 [cos(A/2)]^2=1-[sin(A/2)]^2=2/3 tan^2(B+C)/2 =tan^2(90-A/2)=cot^2A/2 =[cos(A/2)]^2/[sin(A/2)]^2 =2 鎵浠ュ師寮=2+1/3=7/3 ...
绛旓細锛1锛塵=(2sinB,鏍瑰彿3),n=(cos2B,cosB)m//n锛屽垯2sinB*cosB-(鏍瑰彿3)cos2B=0 鍗硈in2B-(鏍瑰彿3)cos2B=0 鍗2sin(2B-60)=0 鎵浠2B=60,B=30搴︺傦紙2锛涓夎褰闈㈢Н S=(1/2)acsinB=1/4ac<=1/8(a^2+c^2)鍏朵腑绛夊彿鎴愮珛鐨勫厖瑕佹潯浠舵槸a=c;鍙︿竴鏂归潰b=鏍瑰彿3-1锛氱敱浣欏鸡瀹氱悊锛堟牴鍙3-1...
绛旓細鍙互鐢ㄥ弽璇佹硶锛屽姞涓婅骞冲垎绾垮畾鐞嗭紝鍜岃竟瑙掑ぇ灏忕殑涓嶇瓑寮忥紙杈瑰ぇ鍒欒澶э級锛屾棤闇娣诲姞浠讳綍杈呭姪绾垮嵆鍙瘉鏄庛傝AB=c锛孊C=a锛CA=b锛岀敱瑙掑钩鍒嗙嚎瀹氱悊锛堝弬鑰冮摼鎺ラ噷鏈夛級锛孊D/DC=AB/AC=c/b锛屼笖BD+DC=BC=a锛屽緱 BD=ac/(b+c)锛孌C=ab/(b+c)锛孊E=1/2AB=c/2锛孋F=b/2銆傚亣璁綛C<1/2(AB+AC)锛...
绛旓細1,sin(A-B)=-cos(A+B)sinAcosB-cosAsinB=sinAsinB-cosAcosB cosB(sinA+cosB)=sinB(sinA+cosB)cosB=sinB tanB=1 B=45 2,b^2=a^2+c^2-2ac*cos45 10=18+c^2-6c c^2-6c+8=0 (c-2)(c-4)=0 c1=2,c2=4
绛旓細锛a+b+c)(a+b-c)=3ab,(a+b)^2-c^2=3ab a^2+b^2-c^2=ab 鐢变綑寮﹀畾鐞嗗緱锛歝osC=(a^2+b^2-c^2)/2ab=1/2 C=60搴
绛旓細鏍规嵁浜屽嶈鍏紡鍙彉褰负:sin2b-鏍瑰彿3cos2b=0 鍗 鏍瑰彿2sin(2b-蟺/6)=0 鎵浠2b-蟺/3=k蟺 (k灞炰簬z)b=蟺/6+k蟺/2鍙堝洜涓篵涓閿愯涓夎褰鐨勫唴瑙,鎵浠=0,b=蟺/6 鍗宠b=30搴.(浜)鍙敤姝ｅ鸡瀹氱悊 a/sina=b/sinb a=bsina/sinb=bsin(180掳-b-c)/sinb=bsin(b+c)/sinb s(鈻abc)=...
绛旓細鎹綑寮﹀畾鐞哸²=b²+c²-2bccosA,寰梑²+c²-a²=2bccosA,浠ュ強闈㈢Н鍏紡S=(1/2)bcsinA.锛屽凡鐭鏉′欢S=(鈭3/4)(b²+c²-a²)鍖栦负 (1/2)bcsinA=(鈭3/4)*2bccosA,灏辨槸tanA=鈭3,閭ｄ箞A=60掳 .銆
绛旓細鏍规嵁浜屽嶈鍏紡鍙彉褰负:sin2B-鏍瑰彿3cos2B=0 鍗 鏍瑰彿2sin(2B-蟺/6)=0 鎵浠2B-蟺/3=k蟺 (k灞炰簬Z) B=蟺/6+k蟺/2鍙堝洜涓築涓閿愯涓夎褰鐨勫唴瑙,鎵浠=0,B=蟺/6 鍗宠B=30搴.(浜)鍙敤姝ｅ鸡瀹氱悊 a/sinA=b/sinB a=bsinA/sinB=bsin(180掳-B-C)/sinB=bsin(B+C)/sinB S(鈻ABC)...
绛旓細(a-c)(sinA+sinC)=(a-b)sinB.鏍规嵁姝ｅ鸡瀹氱悊 (a-c)(a+c)=(a-b)b a²-c²=ab-b²a²+b²-c²=ab 鈭碿osC=(a²+b²-c²)/(2ab)=ab/(2ab)=1/2 鈭礐涓洪攼瑙 鈭碈=蟺/3 鈭碆=蟺-A-C=2蟺/3-A 鈭礏涓閿愯锛鍒2蟺/3-A<...
绛旓細sinC锛屽張A=60掳锛屸埓B+C=120掳锛屽嵆C=120掳锕锛屸埓 =4 锛坰inB+sin120掳cosB锕os120掳sinB锛=4 锛坰inB+ cosB+ sinB锛=6 sinB+6cosB=12锛 sinB+ cosB锛=12sin锛圔+30掳锛夛紝鈭碘柍ABC涓閿愯涓夎褰紝鈭碆鈭堬紙30掳锛90掳锛夛紝鈭碆+30掳鈭堬紙60掳锛120掳锛夆埓 锛屸埓 锛
