If #sin A = 3/5#, what is #sec A#?

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Answer 1 · 2014-11-24T19:38:51

Let #{(mathbf{A} ... "Adjacent"),(mathbf{H} ... "Hypotenuse"),(mathbf{O} ... "Opposite"):}#

Given:
#sin A=3/5={mathbf{O}}/{mathbf{H}} => " Let " {(mathbf{O}=3),(mathbf{H}=5):}#

By Pythagorean Theorem,

#mathbf{A}=sqrt{mathbf{H}^2-mathbf{O}^2}=sqrt{5^2-3^2}=sqrt{16}=4#

Hence,

#sec A=mathbf{H}/mathbf{A}=5/4#

I hope that this was helpful.