BloomLocke367
  • BloomLocke367
http://prntscr.com/7bvs3w @Nnesha
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Nnesha
  • Nnesha
draw a triangle and find 3rd side
BloomLocke367
  • BloomLocke367
what do you mean? (sorry, I really don't understand trig)
Nnesha
  • Nnesha
|dw:1433170800461:dw| given order pair (x,y) x represent cosine y represent sin and \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

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BloomLocke367
  • BloomLocke367
how do I find csc and sec?
Nnesha
  • Nnesha
first you have to find 3rd side of right triangle |dw:1433171058392:dw| sin = opposite /hyp and cos = adjacent /hyp
Nnesha
  • Nnesha
now apply Pythagorean theorem to find 3rd side
BloomLocke367
  • BloomLocke367
13
Nnesha
  • Nnesha
yep right |dw:1433171191619:dw| now what is reciprocal of sec and csc ?? for example tan = sin/cos sec= ?
BloomLocke367
  • BloomLocke367
I don't know
Nnesha
  • Nnesha
mhm that's important part of trig you have to know this \[\huge\rm sec = \frac{ 1 }{ \cos } , \csc =\frac{ 1 }{ \sin } ,\cot =\frac{ \cos }{ \sin } \]
Nnesha
  • Nnesha
cos is adjacent over hyp so 1/adjacent/hpy = ? what is adjacent of right triangle
BloomLocke367
  • BloomLocke367
12
BloomLocke367
  • BloomLocke367
so 1/12/13?
Nnesha
  • Nnesha
yep right now multiply top wit the reciprocal of the bottom
BloomLocke367
  • BloomLocke367
hang on, I have to volunteer now, I'll be back in 2 hours, approximately.
Nnesha
  • Nnesha
aright :-)
BloomLocke367
  • BloomLocke367
@Nnesha I'm back.. I know it took forever, but I never got the chance after I finished working
BloomLocke367
  • BloomLocke367
I got 1? is that right? @Nnesha
Nnesha
  • Nnesha
nope how did you get 1?
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @BloomLocke367 so 1/12/13? \(\color{blue}{\text{End of Quote}}\) multiply top wit the reciprocal of the bottom fraction
BloomLocke367
  • BloomLocke367
ohh I didn't use the reciprocal
BloomLocke367
  • BloomLocke367
I got 169/144
Nnesha
  • Nnesha
mhm how did you get 169 and 133 don't square them |dw:1433196171493:dw|
BloomLocke367
  • BloomLocke367
what?
Nnesha
  • Nnesha
5,12 aand 13 are sides of right triangle what is cos = ? sin = ?
Nnesha
  • Nnesha
\[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]
BloomLocke367
  • BloomLocke367
cos is 12/13 and sin is 5/13
Nnesha
  • Nnesha
yes right now you need sec and csc so sec= 1/cos which is \[\frac{ 1 }{ \frac{12}{13}}\]
Nnesha
  • Nnesha
multiply 1 by the reciprocal of the bottom fraction
Nnesha
  • Nnesha
for example \[\huge\rm \frac{ \frac{ a }{ b } }{ \frac{ c }{ d } } = \frac{ a }{ b } \times \frac{ d }{ c }\]
BloomLocke367
  • BloomLocke367
so 13/12
Nnesha
  • Nnesha
yeah!
Nnesha
  • Nnesha
csc = ?
BloomLocke367
  • BloomLocke367
and that's sec? that's not an option...
Nnesha
  • Nnesha
i did it wrong |dw:1433196991226:dw| sin is opposite so (cosine , sin) x coordinate would be adjacent which suppose to be 5
Nnesha
  • Nnesha
so cos = 5/13 sin = 12/13
Nnesha
  • Nnesha
so csc = what ? and sec = ?
BloomLocke367
  • BloomLocke367
sec=13/5 and csc=13/12
BloomLocke367
  • BloomLocke367
Thank you so much!
Nnesha
  • Nnesha
np :-)
BloomLocke367
  • BloomLocke367
Can you help more?
Nnesha
  • Nnesha
mhm maybe...

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