Trouble with a trig problem

Started by Debster
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Debster

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I've have been trying to figure out how sec45 equals square root 2 for like an hour now. When I solve it like the other trig functions, it comes out 2/square root 2. What am I doing wrong?
 
cosine = adj/hyp
sec = hyp/adj

On a 45/45/90 triangle you have the proportions 1/1/sqrt 2 for the length of the sides. The hyp having the sqrt 2. Choose a side with 45 degrees and take the hyp/adj which will lead to (sqrt 2/1) or just the sqrt 2.

Hope that helped.
 
It did. Thank you!



cosine = adj/hyp
sec = hyp/adj

On a 45/45/90 triangle you have the proportions 1/1/sqrt 2 for the length of the sides. The hyp having the sqrt 2. Choose a side with 45 degrees and take the hyp/adj which will lead to (sqrt 2/1) or just the sqrt 2.

Hope that helped.