How do you convert #(sqrt(3), 1)# into polar forms?

1 Answer
Jan 9, 2016

If #(a,b)# is a are the coordinates of a point in Cartesian Plane, #u# is its magnitude and #alpha# is its angle then #(a,b)# in Polar Form is written as #(u,alpha)#.
Magnitude of a cartesian coordinates #(a,b)# is given by#sqrt(a^2+b^2)# and its angle is given by #tan^-1(b/a)#

Let #r# be the magnitude of #(sqrt3,1)# and #theta# be its angle.
Magnitude of #(sqrt3,1)=sqrt((sqrt3)^2+1^2)=sqrt(3+1)=sqrt4=2=r#
Angle of #(sqrt3,1)=Tan^-1(1/sqrt3)=pi/6#

#implies# Angle of #(sqrt3,1)=pi/6=theta#

#implies (sqrt3,1)=(r,theta)=(2,pi/6)#
#implies (sqrt3,1)=(2,pi/6)#
Note that the angle is given in radian measure.