YLL討論網

This is your work.

(1) let AB=AC=1
then BC=BD=2cos48°

(2) let x=2cos12°, y=2cos36°
y=xxx-3x
2cos24°=xx-2
2cos48°=(xx-2)(xx-2)-2=xxxx-4xx+2
0=2cos(180-72)°+2cos72°=yyy-3y+yy-2=(yy-y-1)(y+2);
y <> -1, so, 0=yy-y-1
=(xxx-3x)^2-(xxx-3x)-1
=x^6-6x^4-xxx+9xx+3x-1
=(xx-x-1)(xxxx+xxx-4xx-4x+1)
but xx-x-1=0之根為2cos36° ,2cos72°, not 2cos12°,
so, (xxxx+xxx-4xx-4x+1)=0
BD=2cos48°=xxxx-4xx+2=1+4x-xxx
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(3)
AD^2 =AB^2+BD^2-2AB*BD*cos12°
=1+(1+4x-xxx)^2-1*(1+4x-xxx)*x
=1+(1+16xx+xxxxxx-2xxx-8xxxx+8x)-x-4xx+xxxx
=xxxxxx-7xxxx-2xxx+12xx+7x+2
=(x^6-6x^4-xxx+9xx+3x-1)-(xxxx+xxx-4xx-4xx+1)+(4-xx)
=4-xx

sin(∠ADB)=AB*sin12°/AD
=sqrt(1-xx/4)/sqrt(4-xx)=1/2
∠ADB=30°

But it is not good to do this type of Q

Raceleader的解
http://yll.loxa.edu.tw/phpBB2/viewtopic.php?t=3186

ABCD是一個四邊形，∠ADB=12°，∠ACB=24°，∠BDC=36°，∠ACD=48°，求∠ABD=？ (只可用解析幾何及三角)