Question

in triunghiul dreptunghic ABC cu m(A)=90°. AC=8 iar sin B=4/5. Bc=?

Answer 1

AC^2=BC^2-AB^2=400-256=144 ⇒ AC=12 CM
SIN C=AB/BC=16/20=4/5
COS C=AC/BC=12/20=6/10=3/5
TG B=AC/AB=12/16
CTG B=AB/AC=16/12
SIN X= CATETA OPUSA/IPOTENUZA
COS X=CATETA ALATURATA/IPOTENUZA
TG X=CATETA OPUSA/CATETA ALATURATA
CTG X=CATETA ALATURATA/CATETA OPUSA=1/TG X
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Answer 2

$sinB=\dfrac{AC}{BC} = \dfrac{4}{5} \Longrightarrow \dfrac{8}{BC}=\dfrac{4}{5} \ \Longrightarrow BC=\dfrac{5\cdot8}{4} = 10\ cm$