a) Ta có:
$\widehat{A} = 90^o$
$\widehat{B} = 58^o$
$\Rightarrow \widehat{C} = 90^o - \widehat{B} = 90^o - 58^o = 32^o$
$\sin\widehat{B} = \dfrac{AC}{BC}$
$\Rightarrow AC = BC.\sin B = 72.\sin58^o \approx 61,06 \, cm$
$\cos\widehat{B} = \dfrac{AB}{BC}$
$\Rightarrow AB = BC.\cos C = 72.\cos58^o \approx 31,15\, cm$
b) Ta có:
$\widehat{A} = 90^o$
$\widehat{B} = 48^o$
$\Rightarrow \widehat{C} = 90^o - \widehat{B} = 90^o - 48^o = 42^o$
$\tan\widehat{B} = \dfrac{AC}{AB}$
$\Rightarrow AB = \dfrac{AC}{\tan B} = \dfrac{20}{\tan48^o} \approx 18,01\, cm$
$\sin\widehat{B} = \dfrac{AC}{BC}$
$\Rightarrow BC = \dfrac{AC}{\sin B} = \dfrac{20}{\sin48^o} \approx 26,91\, cm$
