Đáp án: $AB\approx 2,9\,\left( cm \right)$
Giải thích các bước giải:
$\Delta CDA$ vuông tại $C$
$\Rightarrow \tan \widehat{CDA}=\dfrac{AC}{CD}$
$\Rightarrow AC=CD.\tan \widehat{CDA}$
$\Rightarrow AC=3,2.\tan {{40}^{{}^\circ }}$
$\Delta CDB$ vuông tại $C$
$\Rightarrow \tan \widehat{CDB}=\dfrac{BC}{CD}$
$\Rightarrow BC=CD.\tan \widehat{CDB}$
$\Rightarrow \tan \widehat{CDB}=\dfrac{BC}{CD}\,\,\Rightarrow \,\,BC=CD.\tan \widehat{CDB}=3,2\,.\,\tan 60{}^\circ$
$\Rightarrow BC-AC=3,2\,.\,\tan 60{}^\circ \,-\,3,2\,.\,\tan 40{}^\circ$
$\Rightarrow AB\approx 2,9\,\left( cm \right)$
