www.fltk.net > △ABC在内角A、B、C的对边分别为A,B,C,已知A=BCosC+CsinB.(1)求B;(2)若...

△ABC在内角A、B、C的对边分别为A,B,C,已知A=BCosC+CsinB.(1)求B;(2)若...

作a边上的高线(1)∵a=bcosC+csinB 又a=bcosC+ccosB ∴sinB=cosB∴B=45°(2)∵b=a+c-2accosB∴a+c-√2ac=4≥2ac-√2ac∴ac≤4/(2-√2)=4+2√2即ac最大值为4+2√2∴S△ABC=1/2acsinB≤1/2*(4+2√2)*√2/2=√2+1即△ABC面积的最大值为√2=1

(1)利用正弦定理:a/sinA=b/sinB=c/sinC∵ a=bcosC+csinB∴ sinA=sinBcosC+sinCsinB∵ sinA=sin[π-(B+C)]=sin(B+C)∴ sinBcosC+cosCsinB=sinBcosC+sinCsinB∴ cosCsinB=sinCsinB∴ tanB=1∴ B=π/4(2)S=(1/2)acsinB=(√2/4)ac利用余弦定理4=a+c-2ac*cos(π/4)∴ 4=a+c-√2ac≥2ac-√2ac∴ ac≤4/(2+√2)=2(2+√2)当且仅当a=c时等号成立∴ S的最大值是(√2/4)*2*(2+√2)=√2+1

作a边上的高,则a=bcosC+ccosB∵a=bcosC+csinB∴sinB=cosB∴B=45°

(Ⅰ)由已知及正弦定理得:sinA=sinBcosC+sinBsinC①,∵sinA=sin(B+C)=sinBcosC+cosBsinC②,∴sinB=cosB,即tanB=1,∵B为三角形的内角,∴B=π4;(Ⅱ)S△ABC=12

作a边上的高,则a=bcosC+ccosB∵a=bcosC+csinB∴sinB=cosB∴B=45°∵b^2=a^2+c^2-2accosB∴a^2+c^2-√2ac=4≥2ac-√2ac∴ac≤4/(2-√2)=4+2√2ac最大值为4+2√2∴SABC=1/2acsinB≤1/2*(4+2√2)*√2/2=√2+1∴三角形ABC面积的最大值为√2+1

(1) B = (2) +1 (1)由已知及正弦定理,得sin A =sin B cos C +sin C sin B ,①又 A =π-( B + C ),故sin A =sin( B + C )=sin B cos C +cos B sin C .②由①,②和 C ∈

(1)过A做AD⊥BC与D则BC=bcosC+csinB=CD+ADAD=BC-CD=BD∴△ABD是等腰直角三角形∴∠B=45°(2)问题就是求最大值,肯定跟不等式有关联啦,没有办法彻底回避.CD=2cosCAD=2sinCBC=2sinC+2cosCS=AD*BC/2=2sinC(sinC+cosC)=2sinC+2sinCcosC=1-cos2C+sin2C=√2sin(2C+3π/4)+1≤√2+1因此最大值是√2+1当C=3π/8时取得最大值.

解答:(1)利用正弦定理:a/sinA=b/sinB=c/sinC∵ a=bcosC+csinB∴ sinA=sinBcosC+sinCsinB∵ sinA=sin[π-(B+C)]=sin(B+C)∴ sinBcosC+cosCsinB=sinBcosC+sinCsinB∴ cosCsinB=sinCsinB∴ tanB=1∴ B=π/4(2)S=(1/2)acsinB=(√2/4)ac利用余弦定理4=a+c-2ac*cos(π/4)∴ 4=a+c-√2ac≥2ac-√2ac∴ ac≤4/(2+√2)=2(2+√2)当且仅当a=c时等号成立∴ S的最大值是(√2/4)*2*(2+√2)=√2+1

(1)利用正弦定理化简a=bcosC+csinB,得:sinA=sinBcosC+sinCsinB,由sinA=sin(B+C)=sinBcosC+cosBsinC,代入得:sinBcosC+cosBsinC=sinBcosC+sinCsinB,整

(1)由正弦定理得到:sinA=sinCsinB+sinBcosC,∵在△ABC中,sinA=sin[π-(B+C)]=sin(B+C),∴sin(B+C)=sinBcosC+cosBsinC=sinCsinB+sinBcosC,∴cosBsinC=sinCsinB

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