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

- anonymous

- katieb

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- Nnesha

|dw:1442457266673:dw|
remember this CAST rule for signs

- 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

okay

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## More answers

- anonymous

I'm still so confused haha

- 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

the second one right?

- Nnesha

sin is positive at 2nd quadrant but the given sin is negative |dw:1442457589135:dw|

- 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

|dw:1442457687986:dw|

- Nnesha

in which quadrant sin is negative ?

- anonymous

3 and 4 but tan is less than zero so I'm assuming that it would be quadrant 4

- 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

sin = opposite over hyp
\[\huge\rm sin\theta=\frac{ opposite }{ hyp }=-\frac{ 2 }{ 9 }\]

- prizzyjade

Tan must be in quadrant 4 since, sine is negative.and tangent is less 0 which means it is negative.

- Nnesha

|dw:1442457929676:dw|
remember adjacent and opposite sides depends on the angle

- anonymous

right, so don't I use pythagorean theorem next?

- Nnesha

yes right use pyth theorem to find 3rd side of right triangle

- anonymous

ok, i got the square root of 77. now what?

- Nnesha

|dw:1442458115705:dw|
do you know the definition of sec ?

- Nnesha

csc =1/sin
sec = ??

- anonymous

yeah it would be r/x right?

- Nnesha

hm csc is reciprocal of 1/sin
do you know what's reciprocal of sec ?

- anonymous

huh? lol

- Nnesha

\[\large\rm sin = \frac{ 1 }{ \csc }~~~~\cos=\frac{ 1 }{ \sec } ~~~~\tan=\frac{ 1 }{ \cot }\]

- Nnesha

csc ,sec ,cot are reciprocal of sin ,cos ,tan

- anonymous

ohhh ok so now what do i do?

- Nnesha

to find sec you should know cos
cos = adjacent over hyp so look at the triangle adjacent over hyp = ?

- anonymous

so sec of theta would be 9 over square root 77?

- 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

ohhh ok sooooo ok... that would be 1.025..?

- Nnesha

hmm leave it as a fraction

- 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

just multiply the numerator by numerator and denominator by denominator
do you mean sqrt{77} or just sqrt{7} ??

- anonymous

sqrt(77) and yeah I know that haha, but I just don't know what that would come out to be

- Nnesha

yes that's right

- anonymous

oh great! so that's the answer?

- Nnesha

yes i think so :=)

- 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

and find cos theta lol I just don't know how the 90 less than 0 less than 180 works in the problem

- Nnesha

i've to go sorry one mint left :D

- anonymous

oh lol well thanks anyway!

- Nnesha

|dw:1442459319473:dw|

- Nnesha

are you sure it's 90 less than 0 less than 180 ?

- anonymous

oh my goodness thank you that is so helpful and yes

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