• November 22nd 2006, 07:39 AM
kcsteven
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
• November 22nd 2006, 10:23 AM
Soroban
Hello, Keith!

Your arithmetic is a bit off . . .

Quote:

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.

• November 22nd 2006, 08:13 PM
qbkr21
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.

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

$\text{You Must Know That}$

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

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.

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

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 $Tan(x)$= $\frac{\sqrt{11}}{\sqrt{5}}$

$Tan(x)$= $-\frac{\sqrt{11}}{\sqrt{5}}$