How do you convert #r*(csctheta) = sqrt(11)# into rectangular form? Trigonometry The Polar System Converting Between Systems 1 Answer Shwetank Mauria Feb 23, 2016 #x^2+y^2-sqrt11y=0# Explanation: To convert #r*cscθ=sqrt11# into rectangular form #(x,y)#, we should use the relation between #(x,y)# and #(r,theta)#, which are #r^2=x^2+y^2# and #rcostheta=x# and #rsintheta=y# Note that #csctheta=1/sintheta=r/y#. Hence #r*cscθ=sqrt11# can be written as #r*r/y=sqrt11# or #r^2=sqrt11*y# #x^2+y^2-sqrt11y=0# Answer link Related questions How do you convert rectangular coordinates to polar coordinates? When is it easier to use the polar form of an equation or a rectangular form of an equation? How do you write #r = 4 \cos \theta # into rectangular form? What is the rectangular form of #r = 3 \csc \theta #? What is the polar form of # x^2 + y^2 = 2x#? How do you convert #r \sin^2 \theta =3 \cos \theta# into rectangular form? How do you convert from 300 degrees to radians? How do you convert the polar equation #10 sin(θ)# to the rectangular form? How do you convert the rectangular equation to polar form x=4? How do you find the cartesian graph of #r cos(θ) = 9#? See all questions in Converting Between Systems Impact of this question 1614 views around the world You can reuse this answer Creative Commons License