Question

Cibffhjmgf

求解😭

Answer 1

✨ 最佳解答 ✨

imma_idiot

設 BD=a;CD=b, 利用餘弦定理可得:
9=a^2+b^2-ab (△BCD)
16=(a+2)^2+b^2-(a+2)b (△ABC),相消得 7=4a+4-2b,a=(3+2b)/4;在另∠ACB=C:
16=4+9-12cos(C) (△ABC)=> cos(C)=-1/4,加上-cos(C)=cos(180-C)=cos(∠BCD)=-1/4
a^2=b^2+9+3b/2 (△BCD) =(帶入)=> (9+12b+4b^2)/16=b^2+3b/2+9,12b^2+12b+135=0 => 兩根不是負就是虛 =>無解?

Call me Bravo

That you go, I think to work out b is difficult.

Answer 2

Call me Bravo

For your reference

imma_idiot

m=-tan(60)??

Call me Bravo

You know what I mean, here is m=tan(120),right?

Cibffhjmgf

Thanks