Question

关于三角函数的题,.1、如图,在Rt△ABC中,∠C=90°,点D在BC边上,已知∠ADC=45°,DC=6,sinB=

Answer 1

1

∠ACB=90°,sinB=3/5,sinA=4/5,tanA=4/3,

tan∠BAD=tan(A-45)

=(tanA-1)/(1+tanA)

=(1/3)/(7/3)

=1/7

2

设AC=BC=m

则AB=根号2 m

CD=BD=m/2

BE=BDcos45°=m/(2根号2)

CE^2=BC^2+BE^2-2BC*CEcos45°=m^2+m^2/8-2m*m/(2根号2)*根号2/2=5/8 m^2

CE=根号(5/8) m

CE/sinA=AC/sinAEC

sinAEC=AC/CE *sinA = m/[根号(5/8) m] *根号2/2 = 2根号5 /5