Đáp án: $D$

Giải thích các bước giải:

Kẻ $DH\perp CE$

$\to \widehat{DCH}=180^o-\widehat{DCE}=75^o$

Ta có:

$\cot\widehat{DCH}=\dfrac{HC}{DH}\to HC=DH\cot75^o$

$\cot E=\dfrac{HE}{DH}\to HE=DH\cot45^o$

$\to CE=HE-HC=DH(\cot45^o-\cot75^o)$

$\to 40=DH(\cot45^o-\cot75^o)$

$\to DH=\dfrac{40}{\cot45^o-\cot75^o}$

$\to DH=20(1+\sqrt3)$

Ta có:

$\sin\widehat{DCH}=\dfrac{DH}{CD}$

$\to CD=\dfrac{DH}{\sin\widehat{DCH}}$

$\to CD=\dfrac{20(1+\sqrt3)}{\sin75^o}$

$\to CD= 40\sqrt2$