- Aug 3, 2005
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My trig class was a year ago so I forgot this shiz
Find the rectangular coordinates of the point that has polar coordinates (2, 11pi/6)
So its Polar to Rectangular
11pi/6 is 330 degrees
(2,330)
X= r cos 330
Y= r sin 330
r = square root y^2+x^2
So its 2^2 + 330^2 or 11pi/6^2?
sqrt 330^2 + 2^2 = 2 sqrt 27226
X =-327.1016184
Y= -43.68673994
sqrt 11pi/6^2 + 2^2 = sqrt 4+11pi/36
X=-2.207489256
Y=-0.2948258389
What do I put down for r?
Find the rectangular coordinates of the point that has polar coordinates (2, 11pi/6)
So its Polar to Rectangular
11pi/6 is 330 degrees
(2,330)
X= r cos 330
Y= r sin 330
r = square root y^2+x^2
So its 2^2 + 330^2 or 11pi/6^2?
sqrt 330^2 + 2^2 = 2 sqrt 27226
X =-327.1016184
Y= -43.68673994
sqrt 11pi/6^2 + 2^2 = sqrt 4+11pi/36
X=-2.207489256
Y=-0.2948258389
What do I put down for r?
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