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Math Help - Trig question quadrants

  1. #1
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    Trig question quadrants

    I am working on my homework and I am not sure if I am doing this correctly. Please let me know which one of these are correct, if any.

    Find tan angle in QII, Sin angle=3/4 This is what I tried.
    tan^2 angle +1= Sec^2 angle
    tan^2 angle +1=(1/cos)^2 angle
    (3/4)^2+ 1/cos^2angle=1
    9/16+ 1/cos^2 angle=1
    1/cos^2 angle=7/16
    tan^2 angle=7/16
    tan angle=+- squrt of 7/6, tan angle in Quad II= -squrt of 7/6.

    also this one
    tan^2 angle+1=sec^2 angle
    tan^2 angle+1=(3/4)^2
    tan^2 angle +1=(4/3)^2
    tan^2 angle+1= 16/9
    tan^2 angle= 7/9
    tan angle = +- squrt of 7/9, tan angle = +- squrt of 7/3
    tan angle in quad II= -squrt of 7/3
    I would like to finish my homework but I do not want to do it all wrong. Please let me know if one of these ways is correct, if not could you please explain the right method to me so I can understand and do the rest correctly.

    Thank you very much!
    Keith Stevens
    keithstevens@ctc.net
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  2. #2
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    Hello, Keith!

    Your arithmetic is a bit off . . .


    Given: \sin\theta = \frac{3}{4},\;\theta in Q2. . Find \tan\theta

    Using your approach, I'd do it like this . . .

    . . \cos^2\theta \:=\:1 - \sin^2\theta \;= \; 1 - \left(\frac{3}{4}\right)^2 \:=\:1 - \frac{9}{16} \:=\:\frac{7}{16}

    Then: . \cos\theta \:=\:\pm\frac{7}{4}

    Since \theta is in Q2, cosine is negative: . \cos\theta = -\frac{\sqrt{7}}{4}

    Therefore: . \tan\theta \:=\:\frac{\sin\theta}{\cos\theta} \:=\:\frac{\frac{3}{4}}{-\frac{\sqrt{7}}{4}} \:=\:\boxed{-\frac{3}{\sqrt{7}}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    I prefer to do it like this . . .

    We're given: . \sin\theta = \frac{3}{4},\;\theta\text{ in Q2.}

    Since \sin\theta = \frac{opp}{hyp}, I can make a sketch.
    Code:
          *       |
          : \     |
        3 :   \4  |
          :     \ θ
         -+-------*--------
                  |

    We have: .  opp = 3,\;hyp = 4.

    Using Pythagorus, we find that:  adj = -\sqrt{7}

    . . And now we can write any of the trig functions.

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  3. #3
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    Re:

    Keith \text{It is Andrew, How is Thankgiving?} First, please check out this section of MathhelpFoum, Sorobon does a great job of explaining this concept!!

    http://www.mathhelpforum.com/math-he...-problems.html


    also for this problem you might want to break apart:

    1.

    x) in Q III}" alt="\text{Gx) in Q III}" /> Sin(x)=\frac{3}{4}
    \text{U:Tan(x)}

    \text{You Must Know That}

    (opp)^2+(adj)^2=(hyp)^2

    and since we know that \text{Sin(x)}=\frac{opp}{hyp} you can just plug in the coordinating values and solve:

    Sin(x)=\frac{3}{4} and Tan(x)=\frac{3}{\sqrt{7}}

    You may never have a square root on the bottom so your final answer is:

    \frac{3*\sqrt{7}}{7}


    2.

    x) in Q II}" alt="\text{Gx) in Q II}" /> Sec(x)=\frac{-4}{\sqrt{4}}
    \text{U:Tan(x)}


    Consider the following:

    \text{Tax(x)}= \frac{opp}{adj} so appoach this problem like approached the last one...

    Given that \text{Sec(x)}=\frac{Hyp}{Adj} we now know that \text{Sec(x)}= \frac{-4}{\sqrt{5}}

    substitute your values in for your (opp)^2+({\sqrt{5}^2})=(hyp)^2 and you wil lfind your (opp)={\sqrt{11}}

    Now plug in the value that corrolate with Tan(x) and you will get Tan(x)= \frac{\sqrt{11}}{\sqrt{5}}


    Rember that you are in the 2nd quadrant and Tangent is negative in the 2nd Quadrant so your final answer is:

    Tan(x)= -\frac{\sqrt{11}}{\sqrt{5}}
    Last edited by qbkr21; November 22nd 2006 at 10:56 PM.
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