help with trigonometry, please help, test tomorow:(

I

icat

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Mar 7, 2006
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the problem is sin(theta)=-0.7 for (theta)<=0<=360

i used arcsin(-0.7)=-44.4+360=315.6, which isone of the answers.

for the second answer, i added 270 to get 225.6, which is not the right answer, it is supposed to be 224.4. does anyone know what I am doing wrong, and could somone please explain to me how to solve this problem?
 
also, it is my first time on the forum, have i poste in the right spot?
 
\(\displaystyle \L
| - 44.4 - 180| = 224.4\) which is the correct answer.

Study this carefully.
You know that the answers have reference angles 44.4.
Do you see it?
 
what i mean is, there are many other problems like that, so how do i know when to subtract an angle from theteversuswhen to add an an angle to theta?
 
Think about reference angles.
How do they work?
What is the reference angle of 315.6?
What is the reference angle of 224.4?
 
315.6 would be 44.4, so what i am doing, is basically finding reverse of the reference angle? in which case i would only +/- 360 and 180 to obtain it. so I understand that I would never use 270 in the first place?
 
but then, -44.4-180=-224.4,=360=135.5, whichis a II quadrant answer, and not a third.
 
|arcsin(-0.7)|=44.4, that is the reference angle for what angles in III & IV?
First use absolute value to find what reference angle to use.
Then decide what quadrants we need.
Find the angles with the reference angle in those quadrants.
 
to narrow down my question, why does 224.4 work, and 225.6 does not?
 
ok, thanks, i think i am beginning to get it, so i take 44.4, add 180 to get the III quad answer, and then 360-44.4 for the IV quad answer, so then i only use 180 and 360 to find teh answers right? and never 270 or 90?
 
i understand how to find the 1st quad reference angle for any given angle, its finding an angle that satisfies a certain reference angle that is giving me the triouble.
 
wait where did that post go? did it disappear on its own or is my browser completely php intolerant?
11:28 pm

pka
Senior Member



Joined: 30 Jan 2005
Posts: 1236


At the risk of repeating myself: BECAUSE 44.4 is the reference angle!
Period. That's it.
Now, you may not understand reference angles.
If that is true and you have a test comming, then you have a lot of work to do.
 
Every reference angle is an acute angle, say Θ.
In I it is just Θ.
In II it is 180-Θ.
In III it is 180+Θ.
In VI it is 360-Θ.
 
icat said:
i understand how to find the 1st quad reference angle for any given angle, its finding an angle that satisfies a certain reference angle that is giving me the triouble.
Click to expand...

In one of your previous posts you mentioned using 90 or 270 degrees....you DON't do this.

The reference angle is an angle between 0 and 90 degrees (an acute angle), and it is formed between the terminal side of the given angle and the x-axis. So, if you know the measure of the reference angle in degrees, you'll be adding to or subtracting from 180 degrees or 360 degrees to find the terminal side of the actual angle.

In quadrant II, SUBTRACT the reference angle from 180 degrees.
In quadrant III, ADD the reference angle to 180 degrees.
In quadrant IV, SUBTRACT the reference angle from 360 degrees.

Drawing a diagram should aid you in determining how to use the reference angle.
[/quote]
 
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