The terminal arm of angle #A# goes over the point #(1/5, -1/5)#. What is the exact value of #secA#?

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Answer 1 · 2016-11-08T14:23:12

#secA = sqrt(2)#

The definition of #secA# is #1/cosA = 1/("adjacent"/"hypotenuse") = "hypotenuse"/"adjacent"#.

Now consider the following diagram.

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We know the point on the terminal arm, so we know the two legs of the imaginary triangle above. However, we need the hypotenuse to find #secA#.

#a^2 + b^2 = c^2#

#(1/5)^2 + (-1/5)^2 = c^2#

#1/25 + 1/25 = c^2#

#2/25 = c^2#

#c= sqrt(2)/5 -> "no negative solution since we're talking about the hypotenuse"#

We know that #secA = "hypotenuse"/"adjacent"#, so we have that :

#secA= (sqrt(2)/5)/(1/5)#

#secA = sqrt(2)#

Hopefully this helps!