danaradu70
Tg B=AC/AB=3/4 ⇒ AC=3x, AB=4x aplic teorema lui Pitagora in trg ABC ⇒ BC^2=AB^2+AC^2=(4x)^2+(3x)^2=16x^2+9x^2=25x^2 15^2=225=25x^2 ⇒ x^2=225:25=9 ⇒ x=3 ⇒ AC=3*3 cm=9 cm si AB= 4*3 cm=12 cm sin B=AC/BC=9/15=3/5 cos B=AB/BC=12/15=4/5 ctg B=1/tg B = 4/3
AB=c
AC=b
BC= 15 = a
tg B=
aplic teorema lui Pitagora in trg ABC ⇒ BC^2=AB^2+AC^2=(4x)^2+(3x)^2=16x^2+9x^2=25x^2
15^2=225=25x^2 ⇒ x^2=225:25=9 ⇒ x=3 ⇒ AC=3*3 cm=9 cm si AB= 4*3 cm=12 cm
sin B=AC/BC=9/15=3/5
cos B=AB/BC=12/15=4/5
ctg B=1/tg B = 4/3