在三角形ABC中,内角A.B.C的对边的分别是a.b.c 已知b=acosC+3分之根号3csinA 求A 若a=2求ABC面积的最大值 在三角形abc中.已知a=2，b=2根号2，C=15°，求角...
\u5728\u4e09\u89d2\u5f62ABC\u4e2d,\u5185\u89d2A,B,C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa,b,c,\u5df2\u77e5asin2B=\u6839\u53f73bsina(1)
asin2B=\u221a3bsinA
sinA\u00b72sinBcosB=\u221a3sinBsinA
A\u3001B\u5747\u4e3a\u4e09\u89d2\u5f62\u5185\u89d2\uff0csinA>0\uff0csinB>0
cosB=\u221a3/2
B=\u03c0/6
(2)
sinB=sin(\u03c0/6)=½
sinA=\u221a(1-cos²A)=\u221a(1-⅓²)=2\u221a2/3
sinC=sin(A+B)
=sinAcosB+cosAsinB
=(2\u221a2/3)\u00b7(\u221a3/2)+⅓\u00b7½
=(1+2\u221a6)/6
A=30\u00b0\uff0cB=135\u00b0\uff0cc=\u221a6-\u221a2\u3002
\u89e3\uff1a\u56e0\u4e3acos15\u00b0=cos(45\u00b0-30\u00b0)
=cos45cos30+sin45sin30=(\u221a6+\u221a2)/4
\u90a3\u4e48\u6839\u636e\u4f59\u5f26\u5b9a\u7406\u53ef\u5f97\uff0c
c²=a²+b²-2abcosC
=4+8-8\u221a2*(\u221a6+\u221a2)/4
=(\u221a6-\u221a2)²
\u6240\u4ee5c=\u221a6-\u221a2
\u90a3\u4e48\u6839\u636e\u6b63\u5f26\u5b9a\u7406\uff0ca/sinA=b/sinB=c/sinC\uff0c\u53ef\u5f97\uff0c
2/sinA=(\u221a6-\u221a2)/[(\u221a6-\u221a2)/4]=4\uff0c
\u5219sinA=1/2\uff0c
\u56e0\u4e3aa<b\uff0c\u90a3\u4e48A<B\uff0c\u6240\u4ee5A\u662f\u9510\u89d2\uff0c
\u5219A=30\u00b0\uff0c\u90a3\u4e48B=180-A-C=135\u00b0
\u5373A=30\u00b0\uff0cB=135\u00b0\uff0cc=\u221a6-\u221a2\u3002
\u6269\u5c55\u8d44\u6599\uff1a
1\u3001\u6b63\u5f26\u5b9a\u7406\u6027\u8d28
\u5728\u4efb\u610f\u25b3ABC\u4e2d\uff0c\u89d2A\u3001B\u3001C\u6240\u5bf9\u7684\u8fb9\u957f\u5206\u522b\u4e3aa\u3001b\u3001c\uff0c
\u90a3\u4e48\u6709a/sinA=b/sinB=c/sinC\u3002
2\u3001\u4f59\u5f26\u5b9a\u7406\u6027\u8d28
\u5bf9\u4e8e\u4efb\u610f\u4e09\u89d2\u5f62\uff0c\u4efb\u4f55\u4e00\u8fb9\u7684\u5e73\u65b9\u7b49\u4e8e\u5176\u4ed6\u4e24\u8fb9\u5e73\u65b9\u7684\u548c\u51cf\u53bb\u8fd9\u4e24\u8fb9\u4e0e\u5b83\u4eec\u5939\u89d2\u7684\u4f59\u5f26\u7684\u79ef\u7684\u4e24\u500d\u3002
\u5373\u82e5\u4e09\u8fb9\u4e3aa\uff0cb\uff0cc \u4e09\u89d2\u4e3aA\u3001B\u3001C\uff0c\u90a3\u4e48
c²=a²+b²-2abcosC\u3001b²=a²+c²-2accosB\u3001a²=c²+b²-2cbcosA
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u6b63\u5f26\u5b9a\u7406
b-(a²+b²-c²)/2b=√3/3c×sinA
(2b²-a²-b²+c²)/2b=√3/3c×sinA
(b²+c²-a²)/2b=√3/3c×sinA
(b²+c²-a²)/2bc=√3/3×sinA
cosA=√3/3×sinA
sinA/cosA=3/√3=√3
tanA=√3
∵A是三角形ABC内角
∴A=60°
2、a²=b²+c²-2bccosA
4=b²+c²-bc
b²+c²≥2bc
4≥bc
S=1/2*bc*sinA≤√3
三角形ABC面积最大值=√3
绛旓細B=蟺/6,7,
绛旓細鍦ㄤ笁瑙掑舰涓紝鏈夈愭寮﹀畾鐞嗐戯細asinB=bsinA.鎵浠ワ紝bsinA=鏍瑰彿3acosB锛屽彲浠ュ寲涓 asinB=鏍瑰彿3acosB锛a涓嶆槸0锛屽悓闄や互a锛屽緱鍒 sinB = 鏍瑰彿3 cosB锛屽綋B涓虹洿瑙掓椂锛屽彸杈逛负0锛屽乏杈逛负1锛屼笉绛夈傛墍浠涓嶆槸鐩磋锛宑osB涓嶄负0锛屽悓闄や互cosB寰楀埌 tanB = 鏍瑰彿3. B=60搴︺
绛旓細瑙ｏ細锛1锛夌敱tanA+tanC+tan(蟺/3)=tanAtanCtan(蟺/3) 鍙互寰楀嚭 tanA+tanC=-鈭3*锛1-tanAtanC锛夛紙tanA+tanC锛/锛1-tanAtanC锛=tan(A+C)=-鈭3鍦ㄤ笁瑙掑舰涓 tanB=-tan(A+C)=鈭3 鈭碆=蟺/3 锛2锛夋寮﹀畾鐞哸/sinA=b/sinB=c/sinC鈭达紙a+c锛/(sinA+sinC)=b/sinB=(鈭3/2)/(鈭3/2...
绛旓細cotB=鈭3 鍗矪=30掳 2 鐢变綑寮﹀畾鐞嗙煡 b²=a²+c²-2accosB 鍗 (鈭3)²=3²+c²-2*3ccos30掳 鍗砪²-3鈭3c+6=0 瑙ｅ緱c=2鈭3鎴朿=鈭3 褰揷=2鈭3鏃讹紝S螖ABC=1/2acsinB=1/2*3*2鈭3*sin30掳=3鈭3/2 褰揷=鈭3鏃讹紝S螖ABC=1/2acsinB...
绛旓細1. cosA/2=2鈭5/5 sinA/2=鈭5/5 sinA=2sinA/2*cosA/2=4/5 cosA=(cosA/2)^2-(sinA/2)^2=3/5 AB鍚戦噺鐐逛箻AC鍚戦噺=|AB|*|AC|*cosA=3 |AB|*|AC|=5 S=1/2*|AB|*|AC|*sinA=2 2. b+c=6 |AB|*|AC|=5 bc=5 b=3,c=2鎴朾=2,c=3 浣欏鸡瀹氱悊a^2=b^2+c^2-2bc*...
绛旓細2sin2A=4sinAcosA 鐢眘inC+sin(B-A)=2sin2A寰楋細sinB=2sinA 鍙 sinA/a=sinB/b=sinC/C=鈭3/4 鎵浠 sinA=鈭3/4a锛宻inB=鈭3/4b 鎵浠 b=2a CosC=锛坅^2+b^2-c^2锛/2ab 1/2 =(a^2+4a^2-4)/4a^2 a^2=4/3 涓夎褰鐨勯潰绉疭=1/2absinC=鈭3/ 2a^2 =2鈭3/3 ...
绛旓細ab+b^2=(a+c)(a-c)+(a+c)b ab+b^2=a^2-c^2+ab+bc 鈭碽^2+c^2-a^2=bc 鈶 cosA=(b^2+c^2-a^2)/(2bc)=1/2 鈭碅=60º(2)鍚戦噺AC涓嶤B澶硅涓180º-C 鈭礲=5锛屽悜閲廇C路鍚戦噺CB=5 鈭磡AC||CB|cos(180º-C)=5 鍗-abcosC=5 鈭碼cosC=-1 鏍规嵁...
绛旓細寰梥inB=asinA/4b=5/鈭5.鐢憋紙1锛夌煡锛孉涓洪挐瑙掞紝鍒橞涓洪攼瑙掋傗埓cosB=鈭1-sinB鐨勫钩鏂=2鈭5/5.浜庢槸sin2B=2sinBcosB=4/5 cos2B=1−2sinB鐨勫钩鏂=3/5 鏁卻in锛2B−A锛=sin2BcosA−cos2BsinA=-2鈭5/5.涓夎鍑芥暟鏄暟瀛︿腑灞炰簬鍒濈瓑鍑芥暟涓殑瓒呰秺鍑芥暟鐨勫嚱鏁般傚畠浠殑鏈川鏄换浣曡...
绛旓細鍙堝洜涓恒CD鏄涓夎褰BC鐨涓嚎锛屾墍浠ャ鏄撶煡锛氫笁瑙掑舰BCD鍏ㄧ瓑浜庝笁瑙掑舰AED,鎵浠ャAE=BC=A=8, 瑙扐ED=瑙払CD,鎵浠ャ瑙扐ED+瑙扐CD=瑙払CD+瑙扐CD =瑙扐CB =120搴︼紝鎵浠ャ瑙扖AE=180搴--锛堣AED+瑙扐CD锛=180搴--120搴 锛60搴︺傛墍浠ャ鍦ㄤ笁瑙掑舰ACE涓紝鐢变綑寮﹀畾鐞嗗彲寰楋細CE^2=AC^2+BC^2--2ACxBCxcos...
绛旓細2A-蟺/6=蟺/2 A=蟺/3 B=蟺/3 鈭燗=鈭燘=鈭燙,a=b=c=2 (2)鍓嶉潰鏈敤杩欎釜鏉′欢鐨勬椂鍊,宸茬粡瑙ｅ嚭,鏄瓑杈涓夎褰 涓嶈繃,鐢ㄨ繖涓潯浠跺彲浠ラ獙绠椾竴涓:sin C + sin(B - A)= sin2A sinC=sin(A+B)sinAcosB+cosAsinB+sinBcosA-cosBsinA=2sinAcosA cosAsinB=sinAcosA sinB=sinA 缁撴灉鏄竴鏍风殑....
