I tried taking cos(-3) and sin (-2), but it doesnt work.
2007-12-02 02:06:47 · 6 answers · asked by tchoxi2000 1 in Science & Mathematics ➔ Mathematics
r² = 9 + 4 = 13
r = √13
tan θ = - 2 / 3
θ = 326.3° (4 th quadrant)
2007-12-02 03:04:42 · answer #1 · answered by Como 7 · 3⤊ 1⤋
Let's find r.
r=sqrt(3^2+(-2)^2)
r=sqrt(9+4)
r=sqrt(13)
r=3,6056
x>0, y<0, so 270<θ<360.
tan(θ)=-2/3
Q=326,3099 degrees or 5,6952 radian.
2007-12-02 10:12:05 · answer #2 · answered by Salih D 1 · 1⤊ 1⤋
x= r cosθ and y = r sinθ
now tanθ = y/x
--> tanθ = -2/3
calculate θ from this..
then use r^2 = x^2 + y^2
--> r= (13)^0.5
This will work..
All the best!
2007-12-02 10:12:17 · answer #3 · answered by The learner 2 · 0⤊ 1⤋
r = sqrt(x^2+y^2) = sqrt(13)
theta = tan^-1(-2/3)
2007-12-02 10:11:02 · answer #4 · answered by Mein Hoon Na 7 · 1⤊ 1⤋
r=sqrt[3*3+(-2)*(-2)]=sqrt[13]=3.61
theta=arctan(-2/3)= -33.7deg
2007-12-02 10:12:25 · answer #5 · answered by oldschool 7 · 0⤊ 1⤋
the method of math_kp is best
2007-12-02 10:17:10 · answer #6 · answered by Ahmed Zia 3 · 0⤊ 1⤋