Question 865281
Find the polar coordinates of the point whose 
rectangular coordinates are (4sqrt 3, -4)
Note:: the point is in QIV so the angle is in QIV
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r = sqrt[(4sqrt(3))^2 + 4^2] = sqrt[64] = 8
angle = t = arctan(-4/4sqrt(3)) = arctan(-1/sqrt(3)) = (11/6)pi
Ans: (8,(11/6)pi)
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Find the rectangular coordinates of the point 
whose polar coordinates are (-4, π/6).
x = -4(cos(pi/6)) = -4(sqrt(3)/2) = -2sqrt(3)
y = -4(sin(pi/6)) = -4(1/2) = -2
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Find a rectangular form of the equation 
r = 5 cos θ
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r = sqrt(x^2+y^2)
theta = arctan(y/x)
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Therefore cos(theta) = x/sqrt(x^2+y^2)
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Substituting you get:
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sqrt(x^2+y^2) = 5*x/sqrt(x^2+y^2)
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x^2 + y"^2 = 5x
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Cheers,
Stan H.
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