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A terminal ray passes through (12,-5). Fine the value to all six trig functions

User Sigvardsen
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Okay, here we have this:

Considering the provided angle, we are going to calculate the requested trigonometric functions, so we obtain the following:

So the first thing we will do is calculate the length of the hypotenuse, that is, the distance between the given point and the origin, then we have:


\begin{gathered} r=√((12-0)^2+(-5-0)^2) \\ r=√(12^2+(-5)^2) \\ r=√(144+25) \\ r=√(169) \\ r=13 \end{gathered}

Now we proceed to find the value of each ratio:


sin\beta=(y)/(r)=(-5)/(13)
cos\beta=(x)/(r)=(12)/(13)
tan\beta=(y)/(x)=(-5)/(12)
csc\beta=(r)/(y)=(13)/(-5)=-(13)/(5)
sec\beta=(r)/(x)=(13)/(12)
cot\beta=(x)/(y)=(12)/(-5)=-(12)/(5)

User Cgsd
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