anonymous
  • anonymous
sin theta=-2/9 and tan theta is less than 0. Find sec theta and draw a reference angle. PLEASE HELP IM DYING I don't understand this
Mathematics
katieb
  • katieb
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Nnesha
  • Nnesha
|dw:1442457266673:dw| remember this CAST rule for signs
prizzyjade
  • prizzyjade
Sin theta = -2/9 Tan theta<0 Sec theta = 1/cos theta Tan theta = sintheta/costheta Let tan = -1 -1(costheta)= -2/9 Costheta=2/9 Sectheta=1/(2/9) Sectheta= 9/2
anonymous
  • anonymous
okay

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anonymous
  • anonymous
I'm still so confused haha
Nnesha
  • Nnesha
alright so given sin = -2/9 which is negative 2nd information given is tan < 0 so at which quadrant you should draw right triangle ?
anonymous
  • anonymous
the second one right?
Nnesha
  • Nnesha
sin is positive at 2nd quadrant but the given sin is negative |dw:1442457589135:dw|
anonymous
  • anonymous
OH I was confused which one was the second one, so would it be the fourth one then? the one in the bottom right-hand corner
Nnesha
  • Nnesha
|dw:1442457687986:dw|
Nnesha
  • Nnesha
in which quadrant sin is negative ?
anonymous
  • anonymous
3 and 4 but tan is less than zero so I'm assuming that it would be quadrant 4
Nnesha
  • Nnesha
yes right |dw:1442457799099:dw|draw right triangle in 4th quadrant and remember this \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]
Nnesha
  • Nnesha
sin = opposite over hyp \[\huge\rm sin\theta=\frac{ opposite }{ hyp }=-\frac{ 2 }{ 9 }\]
prizzyjade
  • prizzyjade
Tan must be in quadrant 4 since, sine is negative.and tangent is less 0 which means it is negative.
Nnesha
  • Nnesha
|dw:1442457929676:dw| remember adjacent and opposite sides depends on the angle
anonymous
  • anonymous
right, so don't I use pythagorean theorem next?
Nnesha
  • Nnesha
yes right use pyth theorem to find 3rd side of right triangle
anonymous
  • anonymous
ok, i got the square root of 77. now what?
Nnesha
  • Nnesha
|dw:1442458115705:dw| do you know the definition of sec ?
Nnesha
  • Nnesha
csc =1/sin sec = ??
anonymous
  • anonymous
yeah it would be r/x right?
Nnesha
  • Nnesha
hm csc is reciprocal of 1/sin do you know what's reciprocal of sec ?
anonymous
  • anonymous
huh? lol
Nnesha
  • Nnesha
\[\large\rm sin = \frac{ 1 }{ \csc }~~~~\cos=\frac{ 1 }{ \sec } ~~~~\tan=\frac{ 1 }{ \cot }\]
Nnesha
  • Nnesha
csc ,sec ,cot are reciprocal of sin ,cos ,tan
anonymous
  • anonymous
ohhh ok so now what do i do?
Nnesha
  • Nnesha
to find sec you should know cos cos = adjacent over hyp so look at the triangle adjacent over hyp = ?
anonymous
  • anonymous
so sec of theta would be 9 over square root 77?
Nnesha
  • Nnesha
yes right but you can't leave the radical at the denominator so multiply top and bottom with sqrt{77}\[\frac{ 9 }{ \sqrt{77}} \times \frac{ \sqrt{77} }{ \sqrt{77} }\]
anonymous
  • anonymous
ohhh ok sooooo ok... that would be 1.025..?
Nnesha
  • Nnesha
hmm leave it as a fraction
anonymous
  • anonymous
ummm,... lol ok in that case would it be 9 square root 7 over 77? I don't remember how to simplify those
Nnesha
  • Nnesha
just multiply the numerator by numerator and denominator by denominator do you mean sqrt{77} or just sqrt{7} ??
anonymous
  • anonymous
sqrt(77) and yeah I know that haha, but I just don't know what that would come out to be
Nnesha
  • Nnesha
yes that's right
anonymous
  • anonymous
oh great! so that's the answer?
Nnesha
  • Nnesha
yes i think so :=)
anonymous
  • anonymous
Thank you so so much! would you mind helping me with one more question? It's tan theta = -7/24 and 90 degrees is less than 0 less than 180
anonymous
  • anonymous
and find cos theta lol I just don't know how the 90 less than 0 less than 180 works in the problem
Nnesha
  • Nnesha
i've to go sorry one mint left :D
anonymous
  • anonymous
oh lol well thanks anyway!
Nnesha
  • Nnesha
|dw:1442459319473:dw|
Nnesha
  • Nnesha
are you sure it's 90 less than 0 less than 180 ?
anonymous
  • anonymous
oh my goodness thank you that is so helpful and yes

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