数学求解，初三三角函数！！！！

Question

如题所述

Answer 1

做DE⊥AB于E
∵∠A=∠CBD，∠C=∠AED=90°
∴CD/CB=CB/AC=1/2
令BC=m，则AC=2m，CD=m/2
AD=AC-CD=2m-m/2=3m/2
BD=√(CD²+CB²)=√{(m/2)²+m²} = √5m/2
DE/AD=CD/BD
DE=AD*CD/BD=(3m/2)*(m/2) / (√5m/2) = 3m/(2√5)
sin∠ABD=DE/BD = {3m/(2√5)} / (√5m/2) = 3/5

Answer 2

角打不出来，自己脑补。
1. tgA=BC/AC=1/2, 可以推出sina= 五分之根号五
2.ABD=90-2A sin ABD=sin (90-2A)=cos 2A=1-sinA*sinA=1-4/25= 4/5

Answer 3

角a=30 角b=60
abd=b-cbd=b-a=30
sin abd=0.5

Answer 4

30度