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avatar+230 

Choose the correct answer 

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 Nov 13, 2018
 #1
avatar+4032 
0

\(\text{Given a point }p=(x,y) \\ \text{the angle, }\theta \text{, the vector from }(0,0) \text{ to }p \\ \text{ makes with the positive x-axis is given by}\\ \cos(\theta) = \dfrac{x}{\sqrt{x^2+y^2}}\\ \cos(\theta) = \dfrac{-5}{\sqrt{(-5)^2+(-4)^2}} = \dfrac{-5}{\sqrt{41}} = -\dfrac{5\sqrt{41}}{41}\)

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 Nov 13, 2018

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