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在三角形ABC中,角ABC的对边分别为ABC且满足A/6=B/...

设a/6=b/4=c/3=K, 则a=6K,b=4K,c=3K, ∴ sin A=a/2R=3K/R,sinB=b/2R=2K/R,sinC=c/2R=3K/2R, cosA=(16K^2+9K^2-36K^2)/2(4K*3K)=-11/24, ∴sin2A/(sinB+sinC) =2sinAcosA/(sinB+sinC) =2*3K/R*(-11/24)/(2K/R+3K/2R) =-11/14

2asin(C+π/6) = b+c 根据正弦定理有: 2sinAsin(C+π/6) = sinB+sinC 2sinA{sinCcosπ/6+cosCsinπ/6) = sinB+sinC sinA{√3sinC+cosC) = sinB+sinC √3sinAsinC+sinAcosC = sinB+sinC 又,sinB=sin(A+C) = sinAcosC+cosAsinC ∴ √3sinAsinC+sinAcosC ...

推荐回答中SINB应该等于三分之根号六。所以推荐回答是错误的,大家别看了!

sinA=asinB/b=6*sin(2π/3) / (5√5) = 3√3 / (5√5) = 3√15/25

右边积化和差得 cos2A-cos2B=2cos(π/6-A)cos(π/6+A) cos2A-cos2B=2*1/2[cos(π/6-A+π/6+A)+cos(π/6-A-π/6-A)] cos2A-cos2B=cosπ/3+cos2A cos2B=-1/2 2B=2π/3 B=π/3 2)b=√3且b≤a 2=√3/√3/2=b/sinB=a/sinA=c/sinC=2R a-(1/2*c)=2sinA-1/...

解:(1)3cos(B-C)-1=6cosBcosC, 化简得:3(cosBcosC+sinBsinC)-1=6cosBcosC, 变形得:3(cosBcosC-sinBsinC)=-1, 即cos(B+C)=-1/3 , 则cosA=-cos(B+C)=1/3 ; (2)∵A为三角形的内角,cosA=1/3 , ∴sinA= 根号下(1-cos²A)...

(1) asin2B=√3bsinA sinA·2sinBcosB=√3sinBsinA A、B均为三角形内角,sinA>0,sinB>0 cosB=√3/2 B=π/6 (2) sinB=sin(π/6)=? sinA=√(1-cos2A)=√(1-?2)=2√2/3 sinC=sin(A+B) =sinAcosB+cosAsinB =(2√2/3)·(√3/2)+?·? =(1+2√6)/6

a^2+b^2+c^2+50=6a+8b+10c, a^2+b^2+c^2-6a-8b-10c+50=0, a^2-6a+9+b^2-8b+16+c^2-10c+25=0, (a-3)^2+(b-4)^2+(c-5)^2=0, a=3,b=4,c=5 勾股定理逆定理,判断是直角三角形

先利用余弦定理建立b +c 与a的关系,然后再利用不等式的性质求得范围。

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