What is the polar form of #( 13,-4 )#?

Question

1645 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-18T12:50:56

Polar form is #(13.6 , 5.985) or (13.6 , -0.298) #

#(13 , -4) # lies on #4# th quadrant.

#r =sqrt(13^2 + (-4)^2)= sqrt 185=13.6 #

#tan alpha = 4/13 :. alpha = tan^-1(4/13)=0.298498 #

Since #theta# lies #4# th quadrant.

#theta= 2pi- alpha = 2pi- 0.298498 ~~5.984686# or

#theta= (- alpha)= -0.298498 #

Polar form is #(r, theta) :. (13.6 , 5.985) or (13.6 , -0.298) # [Ans]

Answer 2 · 2017-11-18T13:34:05

#(sqrt(185) , -(19pi)/200 )#

Polar coordinate form is : #( r , theta )#

Cartesian coordinate form #( x , y )#

#x = rcostheta#

#y =rsintheta#

#theta= arctan(y/x)#

#:.#

#13=rcosthetacolor(white)(88)[1]#

#-4=rsinthetacolor(white)(88)[2]# Squaring:

#169=r^2cos^2theta#

#16=r^2sin^2thetacolor(white)(88)# adding [1] and [2]

#185=r^2(sin^2theta+cos^2theta)##color(white)(88888)sin^2theta+cos^2theta=1#

#:.#

#185=r^2=>r=sqrt(185)#

#tantheta=-4/13=>theta=tan^-1(-4/13)~~-0.2985~~-(19pi)/200#

#:.#

#(sqrt(185) , -(19pi)/200 )#