在三角形abc中,sin方a=sin方b sin方c,判断三角形形状
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SINA方=SINB方+SINBSINC+SINC方根据正弦定理a/sinA=b/sinB=c/sinC转化a^2=b^2+c^2+bcbc=-(b^2+c^2-a^2)余弦定理cosA=(b^2+c
a/sinA=2R所以a^2+b^2a^2+b^2所以2abcosC
很简单.根据一个公式sin^2A+sin^2B=1,得出sin^C=1所以角C=90°,所以为直角三角形.
sin^2A+sin^2B=sin^2C利用三角形正弦定理sinA/a=sinB/b=sinC/c显然a^2+b^2=c^2所以边c所对的角C为直角.
设三角形的顶点为A、B、C,对应的边长为a、b、c.过顶点B做AC边上的垂线,设垂线长度为h,则有h=asinC.SΔABC=h*b/2=absinC/2正弦定理a/sinA=b/sinB可得b=as
sin²A+sin²B=2sin²C由正弦定理a^2+b^2=2c^2代入余弦定理:cosC=(a^2+b^2-c^2)/(2ab)=c^2/(2ab)>0所以:cosC
由正弦定理a/sinA=b/sinB=c/sinC=2R,sin²A+sin²B=sin²C两边同乘以4R²得(2RsinA)²+(2RsinB)
用正弦定理化作a^2-b^2+c^2=ac整理得到cosB=a^2-b^2+c^2/2ac=1/2B=π/3
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由正弦定理和已知可以得到:a^2=b^2+c^2.所以三角形为直角三角形.
sin方A+sin方B=sin方C根据正弦定理:a/sinA=b/sinB=c/sinC=2Ra^2/(2R)^2+b^2/(2R)^2=c^2/(2R)^2即:a^2+b^2=c^2,符合勾股定理,
(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B)sin^A*(sin(A+B)-sin(A
sin^2A+sin^2B=sin^2C=sin^2(A+B)=(sinAcosB+sinBcosA)^2=sin^2Acos^2B+sin^2Bcos^2A+2sinAcosAsinBcosB左边减
a²≤b²+c²-bcbc≤b²+c²-a²1/2≤(b²+c²-a²)/2bccosa≥1/2a≤60°
a/sinA=b/sinB=c/sinC=2R=2√2=>a=2RsinA,b=2RsinB,c=2RsinC2√2(sin²A-sin²C)=(a-b)sinB=>4R²
sin^A+sin^B=1sin^A=1-sin^B=con^Bsin^A-cos^B=(sinA+cosB)(sinA-cosB)=0所以sinA=cosB=sin(90-B)或者sinA=-cos
【1】sin方a+sin方b+sin方c=sin方a+sin方b+sin方(180-(a+b))=sin方a+sin方b+sin方(a+b)=sin方a+sin方b+(sina*cosb+cosa*s
选A.因为在三角形ABC中,若sinC=sinA+sinB,又因为sinC=sin(180°-A-B)=sin(A+B)=(sinAcosB+sinBcosA)=sinAcosB+2sinAcosAs
2sinAcosB=sin(A+B)+sin(A-B)=sinC+sin(A-B)=sinC所以sin(A-B)=0所以A=B所以,△ABC是等腰三角形.完毕.
由正弦定理a/sinA=b/sinB=c/sinC=2R,sin²A+sin²B=sin²C两边同乘以4R²得(2RsinA)²+(2RsinB)