## anonymous one year ago Which of the following is true? Tangent is positive in Quadrant I. Sine is negative in Quadrant II. Cosine is positive in Quadrant III. Sine is positive in Quadrant IV.

1. anonymous

@LexiLuvv2431 The true statement in that a tangent is positive in quadrant I.

2. anonymous

tangent is positive in first quadrant

3. Nnesha

|dw:1439494300583:dw| remember the $$\huge\color{green}{\rm CAST }$$ rule

4. anonymous

Thank you guys so much I have a few more can you guys help?

5. Nnesha

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6. Nnesha

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7. anonymous

@LexiLuvv2431 u have ta give me da medal because I answerd it 1 lol

8. anonymous

Find the sine, cosine, and tangent of 120 degrees. and I did lol

9. Nnesha

use unit circle (x,y) where x -coordinate = cos y-coordinate = sin $\rm tan \theta = \frac{ \sin \theta }{ \cos \theta }$

10. Nnesha

|dw:1439494649288:dw|

11. Nnesha

so cos= ? sin=? tan = ?

12. anonymous

Sin is $\frac{ \sqrt{3} }{ 2 }$

13. anonymous

Idk the others though

14. Nnesha

hm yes that's right so if sin =sqrt{3}/2 cos = what look at the same parentheses where x-coordinate represents cos

15. anonymous

waittt okay so then would the cos equal 1/2 and the tan equal $\sqrt{3}$ ?

16. Nnesha

cos is not just 1/2 look at first drawing i posted above cos is negative in which quadrant ?

17. anonymous

18. Nnesha

|dw:1439495263173:dw| quadrant numbers i guess you don't understand why i wrote CAST C= cos S=sin T=tan A= all functions

19. Nnesha

|dw:1439495345327:dw| cos is negative in 2nd and 3rd quadrant

20. Nnesha

now look at the unit circle 120 degree is in 2nd quadrant where cos value should be negative |dw:1439495460867:dw|

21. anonymous

Noo sorry this is soo confusing for me.

22. Nnesha

just focus on it it's not hard you don't have to solve for anything just read the unit circle

23. anonymous

Ohhhh I seee. I get it nowww

24. anonymous

One more similar to this, The point (−3, 1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of θ? Make sure to show all work.

25. Nnesha

|dw:1439495526308:dw| it's like simple (x,y) graph

26. Nnesha

(x,y) x= cos y=sin so what is sin and cos in that equation ?

27. anonymous

-3 and 1??

28. Nnesha

yew which one is cos

29. anonymous

-3

30. Nnesha

yes right so you have draw a right triangle cos =-3 sin =1 sow at which quadrant you should draw a right triangle ?

31. Nnesha

cos = negative (-3) and sin =positive (1) hint!

32. anonymous

33. Nnesha

yep right !!

34. Nnesha

|dw:1439495963007:dw| apply Pythagorean theorem to find 3rd side of right triangle $\huge\rm a^2+b^2=c^2$

35. Nnesha

a=1 b=-3 solve for c

36. anonymous

100?

37. Nnesha

hmm 100 no how did you get that ?

38. Nnesha

replace a with 1 and b with -3 then take square of both numbers and then add them after that take square root

39. anonymous

ohhh I forgot the 3 was negative. So -64

40. Nnesha

show your work how did you get that ?

41. anonymous

I did 1*1 and got one and the -3 * -3 and got -9 and I added that and got -8 and then multiplied -8 * -8 and got -64

42. Nnesha

okay $(1)+(-3)^2=c^2$ when you take even power of negative you will get positive answer

43. Nnesha

and i just noticed you said 100 earlier

44. Nnesha

which is correct but i thought you r saying c =100 c^2=100 take square root both sides you need c not c^2

45. Nnesha

wait no nvm

46. anonymous

Okay now I'm confused

47. Nnesha

$\huge\rm (1)^2 + (-3)^2=c^2$$\rm 1+9=c^2$ (-3 times -3= 9) so now add 1+9$\huge\rm 10=c^2$ now take square root both sides

48. anonymous

10 squared is 100

49. Nnesha

square root not square

50. Nnesha

to cancel out square you should take square root both sides to cancel out square root you should take square both sides so there is c^2 to cancel out square take square root

51. anonymous

Okay I have no idea how to do this part

52. Nnesha

$\huge\rm \sqrt{c^2}= \sqrt{10}$take square root both sides sqrt{c^2} = c so $c =\sqrt{10}$

53. Nnesha

|dw:1439497283978:dw|

54. anonymous

Okayy I see

55. Nnesha

now remember definition of sin,cos and tan you need this $\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }$

56. Nnesha

|dw:1439497367283:dw| opp= opposite side of theta adj= adjacent of theta hyp=hypotenuse of theta

57. anonymous

|dw:1439508247714:dw|

58. anonymous

59. Nnesha

yes right just write value of sin cosine and tan

60. anonymous

How do I do that?

61. Nnesha

$\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }$ just replace opposite , adjacent hyp words with their values

62. Nnesha

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63. anonymous

Thank youu so much

64. Nnesha

like sin = opposite over hyp so opposite side of theta is 1 and hyp = sqrt{10} write it as $\sin \theta =\frac{ 1 }{ \sqrt{10} }$ you can't leave the radical sign at the denominator so multiply top and bottom by square root 10

65. Nnesha

my pleasure:)