Răspuns :
In ΔABC, mA=90, mB=60⇒mC=30⇒conf th 30-60-90 BA=6,BC=12⇒th Pitagora
AC²=BC²-BA²
AC²=144-36=108
AC=6√3
sinC=BA/BC=6/12=1/2
cos C=AC/BC=6√3/12=√3/2
tg C=6/6√3=√3/3
ctgC=6√3/6=√3
AC²=BC²-BA²
AC²=144-36=108
AC=6√3
sinC=BA/BC=6/12=1/2
cos C=AC/BC=6√3/12=√3/2
tg C=6/6√3=√3/3
ctgC=6√3/6=√3
m(C)=90-60=30(grade)
AB=BC/2=6 cm(cateta opusa unghiului de 30 grade este jumatate din ipotenuza)
sinC=sin30=1/2, cos 30=√3/2, tg30=√3/3, ctg30=√3
AB=BC/2=6 cm(cateta opusa unghiului de 30 grade este jumatate din ipotenuza)
sinC=sin30=1/2, cos 30=√3/2, tg30=√3/3, ctg30=√3