a=3 ,b=4, SinA=3/5,则SinC=?
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解决时间 2021-04-01 18:49
- 提问者网友:贪了杯
- 2021-04-01 06:36
a=3 ,b=4, SinA=3/5,则SinC=?
最佳答案
- 五星知识达人网友:毛毛
- 2021-04-01 07:14
sinA =3/5
=>cosA =4/5 or -4/5
b/sinB =a/sinA
sinB = bsinA/a
=4(3/5)/3
= 4/5
=>cosB = 3/5 or -3/5
case 1: cosA =4/5 , cosB =3/5
sinC = sin(A+B)
= sinAcosB + cosAsinB
=(3/5)(3/5) + (4/5)(4/5)
=1
case 2: cosA =-4/5 , cosB =3/5
sinC = sin(A+B)
= sinAcosB + cosAsinB
=(3/5)(3/5) + (-4/5)(4/5)
=-7/25
rejected sinC >0
case 3: cosA =4/5 , cosB =-3/5
sinC = sin(A+B)
= sinAcosB + cosAsinB
=(3/5)(-3/5) + (-4/5)(4/5)
=7/25
case 4: cosA =-4/5 , cosB =-3/5
sinC = sin(A+B)
= sinAcosB + cosAsinB
=(3/5)(-3/5) + (-4/5)(4/5)
=-1 (rejected)
ie
sinC =1 or 7/25
=>cosA =4/5 or -4/5
b/sinB =a/sinA
sinB = bsinA/a
=4(3/5)/3
= 4/5
=>cosB = 3/5 or -3/5
case 1: cosA =4/5 , cosB =3/5
sinC = sin(A+B)
= sinAcosB + cosAsinB
=(3/5)(3/5) + (4/5)(4/5)
=1
case 2: cosA =-4/5 , cosB =3/5
sinC = sin(A+B)
= sinAcosB + cosAsinB
=(3/5)(3/5) + (-4/5)(4/5)
=-7/25
rejected sinC >0
case 3: cosA =4/5 , cosB =-3/5
sinC = sin(A+B)
= sinAcosB + cosAsinB
=(3/5)(-3/5) + (-4/5)(4/5)
=7/25
case 4: cosA =-4/5 , cosB =-3/5
sinC = sin(A+B)
= sinAcosB + cosAsinB
=(3/5)(-3/5) + (-4/5)(4/5)
=-1 (rejected)
ie
sinC =1 or 7/25
全部回答
- 1楼网友:时间的尘埃
- 2021-04-01 08:50
根据正弦定力,可得sinB=4/5
因为b>a,
所以,cosB=3/5或者-3/5
,cosA=4/5
sinC=sin(π-A-B)=sin(A+B)=sinA cosB+cosA sinB
当cosB=3/5 时,sinC=1,此时C=90
当cosB=-3/5时,sinC=7/25
因为b>a,
所以,cosB=3/5或者-3/5
,cosA=4/5
sinC=sin(π-A-B)=sin(A+B)=sinA cosB+cosA sinB
当cosB=3/5 时,sinC=1,此时C=90
当cosB=-3/5时,sinC=7/25
- 2楼网友:蕴藏春秋
- 2021-04-01 07:25
SinA=3/5
cosA=4/5或cosA=-4/5
a^2=c^2+b^2-2bccosA
c^2+16-32/5*c=9
5c^2-32c+35=0
c=5或c=7/5
SinC=c/a*SinA=5/3*3/5=1 或SinC=c/a*SinA=7/5/3*3/5=7/25
1或7/25
cosA=4/5或cosA=-4/5
a^2=c^2+b^2-2bccosA
c^2+16-32/5*c=9
5c^2-32c+35=0
c=5或c=7/5
SinC=c/a*SinA=5/3*3/5=1 或SinC=c/a*SinA=7/5/3*3/5=7/25
1或7/25
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