{急}高中数学题，解三角形
\u9ad8\u4e2d\u6570\u5b66\u9898\uff0c\u89e3\u4e09\u89d2\u5f62h(tan B+tan a)/(tan B-tan a)
\u6211\u7684\u89e3\u6cd5\uff1a
\u5148\u753b\u56fe\uff0c\u8bbe\u89c2\u6d4b\u70b9\u8ddd\u5854\u8ddd\u79bbx\uff0c\u90a3\u4e48\u4ece\u56fe\u4e0a\u590d\u6742\u7684\u5173\u7cfb\u53ef\u5f97\u5854\u7684\u9ad8\u5ea6
H=xtan B-h;H=xtan a+h;
\u89e3\u8fd9\u4e24\u4e2a\u65b9\u7a0b\u2026\u2026\u5f97\u51fa\u7b54\u6848\u2026\u2026
b/a+a/b=(a^2+b^2)/ab=6cosC
\u6839\u636e\u4f59\u5f26\u5b9a\u7406\uff1aa^2+b^2-2abcosC=c^2
=>4ab cosC=c^2
tanC/tanA+tanC/tanB
=tanC[(sinBcosA+sinAcosB)/sinAsinB]
=tanC*sin(A+B)/(sinAsinB)
=tanC*sinC/(sinA*sinB)
=(sinC)^2/(sinA*sinB*cosC)
\u7531\u6b63\u5f26\u5b9a\u7406\uff0c(sinC)^2/(sinAsinB)=c^2/(ab)
\u6240\u4ee5\u4e0a\u5f0f = c^2/(ab cosC)
\u53c8 4ab cosC=c^2
\u6240\u4ee5\u4e0a\u5f0f = 4
又由余弦定理知a^2=b^2+c^2-2bccosA=（b+c）^2-2bc-2bccosA
所以代入上式得S=2bc-2bccosA
又S=1/2bcsinA
所以得2bc-2bccosA=1/2bcsinA 即4-4cosA=sinA
由sinA^2+cosA^=1 j解得sinA=8/17
所以S=1/2bcsinA=4/17*bc≤ 4/17*[(b+c)/2]^2=64/17
所以当且仅当b=c=4时 = 成立 S取得最大值是64/17。
解:S=a^-(b-c)^2
=a^2-b^2-c^2+2bc
=-2bccosA+2bc
=2bc(1-cosA)
而面积公式还有S=(1/2)bcsinA
故sinA=4(1-cosA),解得cosA=15/17,1(舍去)
所以S=2bc(1-15/17)
=4bc/17
≤4[(b+c)/2]^2/17
=4(8/2)^2/17
=64/17
“=”当且仅当b=c=4时取得.
根据余玄定理： a²＝b²＋c²－2bccosA，及b+c=8代入S=a^2-(b-c)^2，化简成只有b与cosA的函数，根据-1≤cosA≤1求解二次函数最大值即可。
a<b+c=8
S=a^2-(b-c)^2<=a^2<64, 当b=c=4时S取最大值,无限接近64.
.....
绛旓細鍦ㄢ柍ABC涓紝瑙扐BC鎵瀵圭殑杈逛簨abc锛屼笖a²+c²-b²=(1/2)ac锛1銆傛眰sin²[(A+C)/2]+cos2B鐨勫硷紝2.鑻=2锛屾眰鈻矨BC闈㈢Н鐨勬渶澶у笺傝В锛1. cosB=(a²+c²-b²)/2ac=[(1/2)ac]/2ac=1/4 sin²[(A+C)/2]+cos2B=sin²[(蟺-B)...
绛旓細瑙ｏ細锛1锛夎AD=x锛屽垯BD=2x锛孉B=x+2x=3x RT鈻矪CD涓 BC²=BD²+CD²=x²+5²=x²+25 鐢变綑寮﹀畾鐞嗗緱锛歝osA=(AD²+AC²-CD²)/(2路AD路AC)cosA=(AB²+AC²-BC²)/(2路AB路AC)(AD²+AC²-CD²)...
绛旓細鈭燙DA=180掳-鈭燚AB=45掳=135掳 AC^2=AD^2+CD^2-2*AD*CDcosCDA=AD^2+CD^2+鈭2*AD*CD BD^2=AD^2+AB^2-2*AD*ABcosDAB=AD^2+AB^2-鈭2*AD*AB AB=CD AC^2*BD^2 =(AD^2+AB^2+鈭2*AD*AB)(AD^2+AB^2-鈭2*AD*AB)=(AD^2+AB^2)^2-(鈭2*AD*AB)^2 =AD^4+...
绛旓細鍒欙細2B+B=180掳 3B=180掳 B=60掳 鐢变綑寮﹀畾鐞嗗緱锛歝osB=(a²+c²-b²)/(2ac)a²+c²-b²=ac 鍥犱负锛2b²=3ac 鍒欙細2a²+2c²-3ac=ac a²-2ac+c²=0 (a-c)²=0 a=c 鎵浠ワ細鈻矨BC鏄瓑杈涓夎褰 鎵浠ワ細A=60掳...
绛旓細sinB(tanA+tanC)=tanAtanC 鍖栫畝寰楋細sinB(sinA/cosA+sinC/cosC)=sinB*sin(A+C)/(cosAcosC)=sinAsinC/(cosAcosC),鍦涓夎褰ABC涓 sinB=sin(A+C),涓婂紡鍙樹负锛歴inB*sinB=sinA*sinC锛岀敱姝ｅ鸡瀹氱悊锛歛/sinA=b/sinB=c/sinC=k锛 浠ｅ叆寰楋細b^2=ac銆傛墍浠 b c 涓変釜杈规垚绛夋瘮鏁板垪 銆傦紙2锛 ...
绛旓細涓夎鍑芥暟 瑕佷箞鑳屽叕寮忥紝璁颁綇鍊笺傝涔堝氨瀛︿細鍏剁殑鎺ㄥ杩囩▼銆傛瘮濡俿in 瀵硅竟姣旀枩杈 cos 閭昏竟姣旀枩杈 tan 瀵硅竟姣旈偦杈 鐒跺悗璁颁綇30掳 60掳鐨勭洿瑙涓夎褰涓夋潯杈归暱搴︿负1锛2锛屾牴鍙3銆45掳瑙掔瓑鑵扮洿瑙掍笁瑙掑舰涓夋潯杈归暱搴︿负1锛1锛屾牴鍙2銆傜劧鍚庤浣忓嶈鍏紡绛夌瓑鍩烘湰鍩虹鐨勪笁瑙掑嚱鏁伴灏遍兘鍙互瑙ｅ喅浜嗐
绛旓細瑙ｏ細鍥燼+c=2b,鏁呯敱姝ｅ鸡瀹氱悊鏈夛細sinA+sinC=2sinB sinA+sinC=2[sin(A+C)/2]cos(A-C)/2= =2[sin(180-B)/2]*cos30(搴︼級=2sin(90-B/2)*鏍瑰彿3/2 =(鏍瑰彿3)cosB/2 鏁咃紝2sinB=(鏍瑰彿3)cosB/2 2*2sinB/2*cosB/2=(鏍瑰彿3)cosB/2,鍥燾osB/2<>0 鏁卻inB/2=(鏍瑰彿3)/4 鍙堬紝...
绛旓細鐜板満鍋氱殑 鏂规硶娌￠敊 鍙兘鏈夌畻閿欏缓璁埆鎶 鐪嬫柟娉 绗竴棰
绛旓細瑙ｄ笁瑙掑舰鍦楂樹腑鏁板蹇呬慨浜旂殑绗竴绔犮傞珮涓暟瀛﹀繀淇簲鐨勭涓绔犳槸瑙ｄ笁瑙掑舰锛涓昏鍐呭鍖呮嫭姝ｅ鸡瀹氱悊銆佷綑寮﹀畾鐞嗐佷笁瑙掑舰鐨勯潰绉叕寮忎互鍙婅В涓夎褰㈢殑搴旂敤涓句緥銆傝В涓夎褰㈡椂锛屽父鐢ㄧ殑鍏紡鍖呮嫭姝ｅ鸡瀹氱悊銆佷綑寮﹀畾鐞嗗拰姝ｅ垏瀹氱悊锛屽畠浠槸瑙ｅ喅涓夎褰㈣竟闀垮拰瑙掑害闂鐨勫熀鏈伐鍏凤紝鍏蜂綋鍏紡濡備笅锛氭寮﹀畾鐞嗭細鍦ㄤ换鎰忎笁瑙掑舰 ABC 涓紝璁...
绛旓細鏍规嵁a/sinA=c/sinC澶勭悊 a=2,c=1 2/sinA=1/sinC sinA=2sinC 鍥犱负0锛渟inA鈮1锛屾墍浠0锛渟inC鈮1/2 褰撹鍦90搴﹀埌180涔嬮棿鏃讹紝姝ｅ鸡鍑芥暟涓哄噺鍑芥暟锛屽嵆瑙掕秺澶у嚱鏁板艰秺灏 涓攕inA=sin(180-A)锛宻inC锛渟inA 鎵浠ヨ嫢C锛90搴︼紝鍒機锛180-A 杩欐牱涓嶈兘鏋勬垚涓夎褰 鍥犳C锛90銆傛鏃舵寮﹀嚱鏁颁负澧炲嚱鏁帮紝sinC...
