## anonymous one year ago 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

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1. Nnesha

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

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

3. anonymous

okay

4. anonymous

I'm still so confused haha

5. 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 ?

6. anonymous

the second one right?

7. Nnesha

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

8. 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

9. Nnesha

|dw:1442457687986:dw|

10. Nnesha

in which quadrant sin is negative ?

11. anonymous

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

12. Nnesha

yes right |dw:1442457799099:dw|draw right triangle in 4th quadrant and remember this $\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }$

13. Nnesha

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

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

15. Nnesha

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

16. anonymous

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

17. Nnesha

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

18. anonymous

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

19. Nnesha

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

20. Nnesha

csc =1/sin sec = ??

21. anonymous

yeah it would be r/x right?

22. Nnesha

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

23. anonymous

huh? lol

24. Nnesha

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

25. Nnesha

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

26. anonymous

ohhh ok so now what do i do?

27. Nnesha

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

28. anonymous

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

29. Nnesha

yes right but you can't leave the radical at the denominator so multiply top and bottom with sqrt{77}$\frac{ 9 }{ \sqrt{77}} \times \frac{ \sqrt{77} }{ \sqrt{77} }$

30. anonymous

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

31. Nnesha

hmm leave it as a fraction

32. anonymous

ummm,... lol ok in that case would it be 9 square root 7 over 77? I don't remember how to simplify those

33. Nnesha

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

34. anonymous

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

35. Nnesha

yes that's right

36. anonymous

oh great! so that's the answer?

37. Nnesha

yes i think so :=)

38. 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

39. anonymous

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

40. Nnesha

i've to go sorry one mint left :D

41. anonymous

oh lol well thanks anyway!

42. Nnesha

|dw:1442459319473:dw|

43. Nnesha

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

44. anonymous

oh my goodness thank you that is so helpful and yes