a) Ta có:
$\tan C = \dfrac{AB}{AC}$
$\Rightarrow AC = \dfrac{AB}{\tan C} = \dfrac{21}{\tan40^o} \approx 25,03 \,cm$
b) Ta có:
$\sin C = \dfrac{AB}{BC}$
$\Rightarrow BC = \dfrac{AB}{\sin C} = \dfrac{21}{\sin40^o} \approx 32,67\, cm$
c) Ta có:
$\widehat{C} = 40^o$
$\Rightarrow \widehat{B} = 50^o$
$\Rightarrow \widehat{ABD} = 25^o$
Ta lại có:
$\cos\widehat{ABD} = \dfrac{AB}{BD}$
$\Rightarrow BD = \dfrac{AB}{\cos\widehat{ABD}} = \dfrac{AB}{\cos25^o} \approx 23,17\, cm$
d) Ta có:
$\tan\widehat{ABD} = \dfrac{AD}{AB}$
$\Rightarrow AD = AB\tan\widehat{ABD} = 21.\tan25^o \approx 9,79\, cm$
$\Rightarrow DC = AC - AD = 25,03 - 9,79 = 15,24 \, cm$
