How do you convert #θ=11pi/6# to rectangular form?

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Answer 1 · 2016-02-22T17:08:47

Let the length of the 'arm' be R then

#x= + R cos(pi/6)" "->" "x=+(R)/2#
#y= -R sin(pi/6)" "->" "y=-(Rsqrt(3))/2" (negative)"#

Given:#" "theta=11/6 pi" "->" "theta" " -=" " 11/6xx180^o = 330^o#
Tont B
Tony B

Given that the triangle in the diagram is 'standardised'

Let R be any length of the arm CP then

#color(blue)("Then for any length "R" by ratio")#

#x=(R)/2" "-> " positive"#

#y=(R)/2xx sqrt(3)" "->" negative "-> -(R)/2xxsqrt(3)#

#color(blue)("Then for any length "R" by trig")#

#x= + R cos(pi/6)" ........ but "cos(pi/6)=1/2#

#y= -R sin(pi/6)" ........ but "sin(pi/6)=sqrt(3)/2#