Results 1 to 3 of 3

Thread: Trig question quadrants

  1. #1
    Junior Member
    Nov 2006

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    May 2006
    Lexington, MA (USA)
    Hello, Keith!

    Your arithmetic is a bit off . . .

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

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

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

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

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

    Therefore: .$\displaystyle \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: .$\displaystyle \sin\theta = \frac{3}{4},\;\theta\text{ in Q2.}$

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

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

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

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

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Oct 2006


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

    also for this problem you might want to break apart:


    $\displaystyle \text{Gx) in Q III}$ $\displaystyle Sin(x)=\frac{3}{4}$
    $\displaystyle \text{U:Tan(x)}$

    $\displaystyle \text{You Must Know That}$

    $\displaystyle (opp)^2+(adj)^2=(hyp)^2$

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

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

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

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


    $\displaystyle \text{Gx) in Q II}$ $\displaystyle Sec(x)=\frac{-4}{\sqrt{4}}$
    $\displaystyle \text{U:Tan(x)}$

    Consider the following:

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

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

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

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

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

    $\displaystyle Tan(x)$=$\displaystyle -\frac{\sqrt{11}}{\sqrt{5}}$
    Last edited by qbkr21; Nov 22nd 2006 at 10:56 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. trig four quadrants
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Dec 4th 2011, 04:16 PM
  2. Circles and Quadrants Question
    Posted in the Geometry Forum
    Replies: 5
    Last Post: Jun 14th 2011, 05:27 AM
  3. trig quadrants
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: May 6th 2010, 02:52 AM
  4. Determining quadrants from pi
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Sep 10th 2009, 10:55 AM
  5. quadrants
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Jun 29th 2009, 04:00 AM

Search Tags

/mathhelpforum @mathhelpforum