Aurreko atalak ikasi eta gero, pres egongo zara hurrengo galderari erantzuteko:Aurreko atalak ikasi eta gero, pres egongo zara hurrengo galderari erantzuteko:
A: Sin30=-x/r ⇨ x=-r.Sin30=-2.1/2=-1
Cos30=-y/r ⇨ y=-r.Cos30=-2.erro3/2=-3
A=(-1,-erro3)
A=-1-erro3i
A ⇨ r=(x)ber2+(y)ber2=(-1)ber2 +(-erro3)ber2=2
x=arctan-erro3/-1=arctanerro3=240gradu
A=2
240gradu
B: Sin60=-y/r ⇨ y=-r.Sin60=-2.erro3/2=-3
Cos60=x/r ⇨ x=r.Cos60=2.1/2=1
B=(1,-erro3)
B=1-erro3i
B ⇨ r=(x)ber2+(y)ber2=(1)ber2 +(-3)ber2=2
x=arctan-erro3/1=arctan-erro3=300gradu
B=2
B: Sin60=-y/r ⇨ y=-r.Sin60=-2.erro3/2=-3
Cos60=x/r ⇨ x=r.Cos60=2.1/2=1
B=(1,-erro3)
B=1-erro3i
B ⇨ r=(x)ber2+(y)ber2=(1)ber2 +(-3)ber2=2
x=arctan-erro3/1=arctan-erro3=300gradu
B=2
300gradu
C=(2,0)
C=2+0i
C ⇨ r=(x)ber2+(y)ber2=(2)ber2 +(0)ber2=2
x=arctan0/2=arctan0=Ogradu
C=2O0
D: Sin30=x/r ⇨ x=r.Sin30=2.1/2=1
Cos30=y/r ⇨ y=r.Cos30=2.erro3/2=erro3
D=(1,erro3)
D=1+erro3i
D ⇨ r=(x)ber2+(y)ber2=(1)ber2 +(3)ber2=2
x=arctanerro3/1=arctanerro3=60gradu
D=2
C=(2,0)
C=2+0i
C ⇨ r=(x)ber2+(y)ber2=(2)ber2 +(0)ber2=2
x=arctan0/2=arctan0=Ogradu
C=2O0
D: Sin30=x/r ⇨ x=r.Sin30=2.1/2=1
Cos30=y/r ⇨ y=r.Cos30=2.erro3/2=erro3
D=(1,erro3)
D=1+erro3i
D ⇨ r=(x)ber2+(y)ber2=(1)ber2 +(3)ber2=2
x=arctanerro3/1=arctanerro3=60gradu
D=2
60gradu
E: Sin30=-x/r ⇨ x=-r.Sin30=-2.1/2=-1
Cos30=y/r ⇨ y=r.Cos30=2.erro3/2=erro3
E=(-1,erro3)
E=-1+erro3i
E ⇨ r=(x)ber2+(y)ber2=(-1)ber2 +(3)ber2=2
x=arctan3-1=arctan-3=120gradu
E=2
E: Sin30=-x/r ⇨ x=-r.Sin30=-2.1/2=-1
Cos30=y/r ⇨ y=r.Cos30=2.erro3/2=erro3
E=(-1,erro3)
E=-1+erro3i
E ⇨ r=(x)ber2+(y)ber2=(-1)ber2 +(3)ber2=2
x=arctan3-1=arctan-3=120gradu
E=2
120gradu
F=(-2,0)
F=-2+0i
F ⇨ r=(x)ber2+(y)ber2=(-2)ber2+(0)ber2=2
x=arctanO/-2=arctan0=180gradu
F=(-2,0)
F=-2+0i
F ⇨ r=(x)ber2+(y)ber2=(-2)ber2+(0)ber2=2
x=arctanO/-2=arctan0=180gradu
F=2
180gradu
Blog baten izenburuan ezin da idatzi mezu bat, hau da, arazoren bat izanez gero, ariketa egiteko, gmail-eko mezuren baten bidez jakinarazi!!
ResponderEliminar