What is the polar form of #( -18,-61 )#?

1 Answer
Jan 14, 2018

#=> color(red)( (sqrt(4045) , tan^(-1) (61/18) + pi ) " In radians" #

#=> color(red)( (sqrt(4045) , tan^(-1) (61/18) + 180^circ) " In degrees" #

Explanation:

We must first use our knowledge of polar coordinates...

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We see from this diagram, that:

#color(red)(x = rcostheta #

#color(red)(y = rsintheta #

#color(red)( r^2 = r^2cos^2x + r^2 sin^2x = x^2 + y^2 " Using pythagerous" #

#=> (x,y) -= (rcostheta , rsintheta ) #

So we can find #r#:

#r^2 = (-18)^2 + (-61)^2 #

#=> r^2 = 324 + 3721 = 4045 #

#=> r = sqrt(4045) #

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We can now find #alpha#

#=> tanalpha = 61/18 #

#=> alpha = tan^(-1) (61/18) #

Now we need the angle form the positive #x# axis

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So hence

#theta =alpha + pi " Radians" #

#theta = alpha + 180^circ " Degrees" #

#=> theta = tan^(-1) ( 61/18) + pi #

or #=> theta = tan^(-1) (61/18) + 180^circ #

#=> color(red)( (sqrt(4045) , tan^(-1) (61/18) + pi ) " In radians" #

#=> color(red)( (sqrt(4045) , tan^(-1) (61/18) + 180^circ) " In degrees" #