3个简单的数学三角函数题!1.在三角形ABC中,sinA=sinB+sinC/cosB+cosC .判断三角形的形状.2.四边形ABCD中,AD垂直于CD,AD=10.AB=14 ∠BDA=60°∠BCD=135°求BC长.3.三角形ABC中,C=2A,a+c=10,cosA=3/4 求b.
2019-04-13
3个简单的数学三角函数题!
1.在三角形ABC中,sinA=sinB+sinC/cosB+cosC .判断三角形的形状.
2.四边形ABCD中,AD垂直于CD,AD=10.AB=14 ∠BDA=60°∠BCD=135°
求BC长.
3.三角形ABC中,C=2A,a+c=10,cosA=3/4 求b.
优质解答
1.
sinA=(sinB+sinC)/(cosB+cosC)
sin(B+C)=(sinB+sinC)/(cosB+cosC)
sinBcosC+cosBsinC=(sinB+sinC)/(cosB+cosC)
sinBcosBcosC+sinB(cosC)^2+(cosB)^2sinC+cosBsinCcosC=sinB+sinC
sinBcosBcosC+cosBsinCcosC=sinB-sinB(cosC)^2+sinC-(cosB)^2sinC
sinBcosBcosC+cosBsinCcosC=sinB(sinC)^2+(sinB)^2sinC
cosBcosC(sinB+sinC)=sinBsinC(sinB+sinC)
(cosBcosC-sinBsinC)(sinB+sinC)=0
cos(B+C)(sinB+sinC)=0
sinB+sinC≠0
所以cos(B+C)=0
B+C=90度,三角形ABC为直角三角形
2.
做BE垂直于AD,交AD于点E 设DE长度为m,
则AE = AD-DE = 10-m
∵∠BDA=60度,
∴BE=m√3
在Rt△ABE中,由AB^2 = AE^2+BE^2
14^2 = (10-m)^2 + 3m^2
∴m = 8
∵∠BCD=135度,
∴BC= m√2
∴BC = 8√2
3.
a/sinA=c/sinC=C/(2sinAcosA)=2c/3sinA3a
=2ca+c
=10
a=4,c=6
a^2=b^2+c^2-2b*c*CosA
16=b^2+36-9*bb=5,(b=4舍去)
[或用C^2=a^2+b^2-2a*b*CosB,36=16+b^2-b,b=5,(b=-4舍去)]
1.
sinA=(sinB+sinC)/(cosB+cosC)
sin(B+C)=(sinB+sinC)/(cosB+cosC)
sinBcosC+cosBsinC=(sinB+sinC)/(cosB+cosC)
sinBcosBcosC+sinB(cosC)^2+(cosB)^2sinC+cosBsinCcosC=sinB+sinC
sinBcosBcosC+cosBsinCcosC=sinB-sinB(cosC)^2+sinC-(cosB)^2sinC
sinBcosBcosC+cosBsinCcosC=sinB(sinC)^2+(sinB)^2sinC
cosBcosC(sinB+sinC)=sinBsinC(sinB+sinC)
(cosBcosC-sinBsinC)(sinB+sinC)=0
cos(B+C)(sinB+sinC)=0
sinB+sinC≠0
所以cos(B+C)=0
B+C=90度,三角形ABC为直角三角形
2.
做BE垂直于AD,交AD于点E 设DE长度为m,
则AE = AD-DE = 10-m
∵∠BDA=60度,
∴BE=m√3
在Rt△ABE中,由AB^2 = AE^2+BE^2
14^2 = (10-m)^2 + 3m^2
∴m = 8
∵∠BCD=135度,
∴BC= m√2
∴BC = 8√2
3.
a/sinA=c/sinC=C/(2sinAcosA)=2c/3sinA3a
=2ca+c
=10
a=4,c=6
a^2=b^2+c^2-2b*c*CosA
16=b^2+36-9*bb=5,(b=4舍去)
[或用C^2=a^2+b^2-2a*b*CosB,36=16+b^2-b,b=5,(b=-4舍去)]