How do you convert #(sqrt3/2 ,1/2 )# into polar coordinates? Precalculus Polar Coordinates Converting Coordinates from Polar to Rectangular 1 Answer sankarankalyanam Jul 9, 2018 #color(red)(r = 1, theta = (pi/6)^c " or " 30^@# Explanation: #(x,y) = (sqrt3/2, 1/2)# #r = sqrt(x^2 + y^2) = sqrt((sqrt3/2)^2 + (1/2)^2) = 1# #theta = arctan ((1/2) / (sqrt3/2)) = (pi/6)^c " or " 30^@# Answer link Related questions What is the formula for converting polar coordinates to rectangular coordinates? How do I convert polar coordinates #(5, 30^circ)# to rectangular coordinates? How do I convert polar coordinates #(3.6, 56.31)# to rectangular coordinates? How do I convert polar coordinates #(10, -pi/4)# to rectangular coordinates? How do I convert polar coordinates #(4,-pi/3)# to rectangular coordinates? How do I convert polar coordinates #(6, 60^circ)# to rectangular coordinates? How do I convert polar coordinates #(-4, 230^circ)# to rectangular coordinates? What is the Cartesian equivalent of polar coordinates #(sqrt97, 66^circ)#? What is the Cartesian equivalent of polar coordinates #(2, pi/6)#? What is the Cartesian equivalent of polar coordinates #(7, pi)#? See all questions in Converting Coordinates from Polar to Rectangular Impact of this question 2803 views around the world You can reuse this answer Creative Commons License