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OpenStudy (fibonaccichick666):
First you need to figure out what A is, any ideas?
OpenStudy (anonymous):
Would it be 300 degrees?
OpenStudy (fibonaccichick666):
Explain how you arrived at that conclusion for me please
OpenStudy (anonymous):
If cos of A is 1/2 and in quadrant 4 on the unit circle shows 300... Am I off on thinking that way?
OpenStudy (fibonaccichick666):
uhm, well let me show you how I think about it. (it is not 300) So in the 1st quad, which angle has a cosine of 1/2?
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OpenStudy (anonymous):
60
OpenStudy (fibonaccichick666):
I apologize, I was incorrect. It is 300, but let's still say why
OpenStudy (fibonaccichick666):
so if we graph,|dw:1395190978468:dw|
OpenStudy (fibonaccichick666):
Can you tell me the value of each side in terms of x?
OpenStudy (fibonaccichick666):
and label however you choose for the angles
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OpenStudy (anonymous):
Not sure I follow
OpenStudy (fibonaccichick666):
so you know how for 45 deg we get a 1, 1, √2 triangle, how about for the one with 60 deg
OpenStudy (anonymous):
x, 2x, x sqrt 3
OpenStudy (fibonaccichick666):
good so can you label those on our picture?
OpenStudy (anonymous):
x= x, h= 2x, y= x sqrt 3
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OpenStudy (fibonaccichick666):
|dw:1395192172888:dw|
And what would be the value of angle z?
OpenStudy (anonymous):
300
OpenStudy (fibonaccichick666):
in regards to the y axis
OpenStudy (anonymous):
30
OpenStudy (fibonaccichick666):
ok so that is how we are going to draw A/2. You have determined that A is 300 now A/2=?
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OpenStudy (anonymous):
150
OpenStudy (fibonaccichick666):
so can you graph that for me please
OpenStudy (anonymous):
I'm on my iPad and it can't access the utilities
OpenStudy (fibonaccichick666):
k draw it for yourself then
OpenStudy (fibonaccichick666):
then tell me which quadrant it falls in and what angle is formed with the y axis
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OpenStudy (anonymous):
QII and 60 degrees
OpenStudy (fibonaccichick666):
|dw:1395193203972:dw|
OpenStudy (fibonaccichick666):
good, so now, can you figure out what the cosine value is?
OpenStudy (anonymous):
-sqrt. 3/2 ?
OpenStudy (fibonaccichick666):
yuppers, and why?
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OpenStudy (anonymous):
Idk because if I use the half angle formula I get sqrt of (3/4)...
OpenStudy (fibonaccichick666):
well... you could use that, but it's a lot easier to just look at the pic. The reference angle will always be the one that is created by the original angle and the x axis
OpenStudy (fibonaccichick666):
|dw:1395194285467:dw|
Think about how we labelled the sides before
OpenStudy (fibonaccichick666):
|dw:1395194307787:dw|
OpenStudy (fibonaccichick666):
use that to come to your conclusion for this new triangle
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OpenStudy (anonymous):
Well my prof wants to see that formula.. Is that answer I got right?
OpenStudy (fibonaccichick666):
oh well, i didn't know that haha but tell me it again? you gave me two answers. Also you should learn this way it's a heck of a lot easier than remembering all of the formulas
OpenStudy (anonymous):
Haha no worries. I got -sqrt of 3/4
OpenStudy (fibonaccichick666):
yup that's it
OpenStudy (fibonaccichick666):
I personally do it this way. I can't remember my name half the time so this def helps
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