How do you convert #(2sqrt3, 2)# into polar form?
1 Answer
Dec 18, 2016
Explanation:
To convert from
#color(blue)"cartesian to polar form"# That is
#(x,y)to(r,theta)#
#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(r=sqrt(x^2+y^2))color(white)(2/2)|)))# and
#color(red)(bar(ul(|color(white)(2/2)color(black)(theta=tan^-1(y/x))color(white)(2/2)|)))#
#"here " x=2sqrt3" and " y=2#
#rArrr=sqrt((2sqrt3)^2+2^2)=sqrt(12+4)=4#
#"Now, " (2sqrt3,2)" is in the first quadrant " # so we must ensure that#theta# is in the first quadrant.
#theta=tan^-1(2/(2sqrt3))=tan^-1(1/sqrt3)=pi/6# and
#pi/6" is the reference angle in the first quadrant"#
#rArr(2sqrt3,2)to(4,pi/6)#
