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Thread: COMPASS trigonometry question

  1. #1
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    COMPASS trigonometry question

    Hi, I'm taking the a Math placement test this Saturday so I've been practicing for that, but here's another problem I just can't figure out:

    if sinθ= 5/13 and tanθ < 0
    then secθ=

    I know that sin = opposite/hypotenuse
    therefore the opposite side is 5 and the hypotenuse is 13

    I also know that sec = hypotenuse/ adjacent

    I don't know how to get the adjacent. I've tried law of sines a/sinalpha = b/sinbeta but I only have on angle (sinθ=5/13 = 22,62 degrees ?)

    Then I tried law of cosines, but same problem here, I don't have the angle opposite the side I'm looking for.

    The answer is supposed to be -13/12.
    Could anyone please give me a hint on how to get there?
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by tami-chan87 View Post
    Hi, I'm taking the a Math placement test this Saturday so I've been practicing for that, but here's another problem I just can't figure out:

    if sinθ= 5/13 and tanθ < 0
    then secθ=

    I know that sin = opposite/hypotenuse
    therefore the opposite side is 5 and the hypotenuse is 13

    I also know that sec = hypotenuse/ adjacent

    I don't know how to get the adjacent. I've tried law of sines a/sinalpha = b/sinbeta but I only have on angle (sinθ=5/13 = 22,62 degrees ?)

    Then I tried law of cosines, but same problem here, I don't have the angle opposite the side I'm looking for.

    The answer is supposed to be -13/12.
    Could anyone please give me a hint on how to get there?
    Pythagoras will give you the value for the other side.
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  3. #3
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    Hello, tami-chan87!

    If $\displaystyle \sin\theta = \frac{5}{13}$ and $\displaystyle \tan\theta < 0$, find $\displaystyle \sec\theta.$

    The sine is positive; $\displaystyle \theta$ is in Quadrant 1 or 2.
    The tangent is negative; $\displaystyle \theta$ is in Quadrant 2 or 4.
    . . Hence: $\displaystyle \theta$ is in Quadrant 2.

    We have: .$\displaystyle \sin\theta \:=\:\frac{5}{13} \:=\:\frac{opp}{hyp} \quad\Rightarrow\quad opp = 5,\;\;hyp = 13 $


    Pythagorus says: .$\displaystyle opp^2 + adj^2 \:=\:hyp^2$

    So we have: .$\displaystyle 5^2 + adj^2 \:=\:13^2\quad\Rightarrow\quad adj^2 \:=\:144\quad\Rightarrow\quad adj \:=\:\pm12$

    Since $\displaystyle \theta$ is in Quadrant 2, $\displaystyle a \:=\:-12$


    Therefore: .$\displaystyle \sec\theta \;=\;\frac{hyp}{adj} \;=\;\frac{13}{\text{-}12} \;=\;\boxed{-\frac{13}{12}}$

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  4. #4
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    thanks

    Thank you so much!
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