How do you convert #(2sqrt3, 2)# into polar form?

1 Answer
Dec 18, 2016

#(4,pi/6)#

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)#