Question

几何题一道如图,△ABC中,∠BAC=80°,∠ABC=60°,D为△ABC内一点,且∠BAD=10°,∠ABD=20°

Answer 1

因为(AD/BD)*(BD/CD)*(CD/AD)=1,设角ACD=x度,

根据正弦定理可得(角度的度省略)(sin20/sin10)*(sin(40-x)/sin40)*(sin70/sinx)=1

而sin20/sin10=2cos10,sin40/sin70=sin40/cos20=2sin20,

所以2cos10sin(40-x)=2sinxsin20,而sin20=2sin10cos10,

所以sin(40-x)=2sinxsin10,又2sinxsin10=cos(x-10)-cos(x+10),

则sin(40-x)=cos(x-10)-cos(x+10)=sin(100-x)-sin(80-x),

那么sin(80-x)=sin(100-x)-sin(40-x)=2cos(70-x)sin30=cos(70-x)=sin(20+x)

则80-x=20+x可得x=30即角ACD=30度