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

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

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

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

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

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

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

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