What are the components of the vector between the origin and the polar coordinate #(3, (13pi)/12)#?

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Answer 1 · 2018-03-14T01:50:42

Coordinates are #x = -3.1058, -0.7765#

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Given : #r = 3, theta = (13pi)/12# since #(13pi)/12#, the vector is in the third quadrant where x & y are negative as cos & sin are negative.

#x = r cos theta = 3 * cos ((13pi)/12) = 3 * (-cos ((pi)/12))#

#x = - 3.1058#

#y = r * sin theta = 3 * sin ((13pi)/12) = 3 * (-sin (pi/12))#

#y = - 0.7765#

Slope #m = tan theta = y / x =( -0.7765) / ( -3.1058) = 0.25#