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Homework Help: Tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2) ?

  1. Jul 13, 2008 #1
    tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2) ???

    Folks,

    In the solution guide to Kline's calculus book, he gives the following in working a problem:

    tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2)

    Can someone explain why this is the case?

    Thanks,

    Tristan
     
  2. jcsd
  3. Jul 13, 2008 #2

    rock.freak667

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    Re: tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2) ???

    tan(t)=Opposite/Adjacent. Draw a right-angled triangle with the side opposite to the angle ,t, as x and the adjacent side to it as a. Then use pythagoras' theorem.
     
  4. Jul 13, 2008 #3
    Re: tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2) ???

    I just figured this out on my own. For anyone who is curious:

    tan(t) = x/a means the opposite and adjacent legs of a right triangle with angle t, are x and a respectively. By the Pythagorean Therom, this implies that the hypotenuse is sqrt(a^2 + x^2). Since sin t = opp/hyp, sin t = x/sqrt( a^2 + x^2). QED
     
  5. Jul 16, 2008 #4

    Gib Z

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    Re: tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2) ???

    lol ?
     
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