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anonymous

  • 5 years ago

Sketch a triangle that has acute angle θ, and find the other five trigonometric ratios of θ. Tan(degree sign)= sqr of 15

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  1. anonymous
    • 5 years ago
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    sin(θ) = cos(θ) = csc(θ) = sec(θ) = cot(θ) =

  2. anonymous
    • 5 years ago
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    I'll take a shot at it..... So tan is sin/cos which is y/x So x = 1 and y = sqrt15 Then do x^2 + y^2 = r^2

  3. anonymous
    • 5 years ago
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    Does that seem right so far?

  4. anonymous
    • 5 years ago
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    I have no idea. I'm been trying to figure this out for a while now. Thanks for trying! If it is right kudos to you :)

  5. anonymous
    • 5 years ago
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    I can keep going XD

  6. anonymous
    • 5 years ago
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    So x^2 + y^2 = r^2 (r is the hypotenuse) (1)^2 + (sqrt15)^2 = r^2 1+ 15=r^2 16=r^2 sqrt16=r r=4

  7. anonymous
    • 5 years ago
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    So now you have all the needed numbers to finish the problem. x = 1 y = sqrt15 r = 4

  8. anonymous
    • 5 years ago
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    sin = sqrt15/4 cos = 1/4 tan = sqrt15 csc = 4/sqrt15 sec = 4 cot = 1/sqrt15

  9. anonymous
    • 5 years ago
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    Thank you so much! But the Sin= sqrt15/4 is not the right answer.

  10. anonymous
    • 5 years ago
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    I can't seem to figure that one out.

  11. anonymous
    • 5 years ago
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    Is everything else right?

  12. anonymous
    • 5 years ago
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    Yes

  13. anonymous
    • 5 years ago
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    Is it (sqrt15)/4 or is the 4 inside the sqrt of 15?

  14. anonymous
    • 5 years ago
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    It is (sqrt15)/4 because it is y/r.

  15. anonymous
    • 5 years ago
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    Ok, yeah that's the right answer. Thanks again :)

  16. anonymous
    • 5 years ago
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    Oh okay, welcome! XD

  17. amistre64
    • 5 years ago
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    you got it?

  18. anonymous
    • 5 years ago
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    Yeah. But I have another problem. Use the figure and the value given below to answer the following questions. Assume A = 4. (a) Express x in terms of trigonometric ratios of θ. (b) Express y in terms of trigonometric ratios of θ.

  19. amistre64
    • 5 years ago
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    ok...gonna need the picture tho :)

  20. amistre64
    • 5 years ago
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    or use this to let me know what I am looking at :)

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  21. anonymous
    • 5 years ago
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    A is the Adjacent angle, y is the hypotenuse, and x is the opposite. Acute triangle.

  22. anonymous
    • 5 years ago
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    Yes, that pic.

  23. amistre64
    • 5 years ago
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    LIKE THIS? opps caps were on lol

  24. amistre64
    • 5 years ago
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  25. anonymous
    • 5 years ago
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    The A is on the bottom, but everything else is in the right place.

  26. amistre64
    • 5 years ago
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    are we dealing with a right triangle?

  27. anonymous
    • 5 years ago
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    yes

  28. amistre64
    • 5 years ago
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    A is bottom left angle?

  29. anonymous
    • 5 years ago
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    Just the bottom angle. I'll try to put up the pic

  30. anonymous
    • 5 years ago
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    Here!

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  31. amistre64
    • 5 years ago
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    ahhh...thats better, when you said A was the adjacent angle it threw me :)

  32. anonymous
    • 5 years ago
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    Sorry :[

  33. amistre64
    • 5 years ago
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    'sok....ill live ;)

  34. amistre64
    • 5 years ago
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    x is equal to "sin(t)" sin(t) = x/y x = y * sin(t)

  35. amistre64
    • 5 years ago
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    probably want my to use A for that eh....

  36. amistre64
    • 5 years ago
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    A = 4...that helps....

  37. anonymous
    • 5 years ago
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    I don't understand how to use it in the equation.

  38. amistre64
    • 5 years ago
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    I think its asking for all the trig functions that use the "x" there and how we would solve for "x": x = yadayadayada stuff like that, might be overkill, but better safe than sorry :)

  39. amistre64
    • 5 years ago
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    sin(t) = x/y ; x = y * sin(t) csc(t) = y/x ; x = y/csc(t) tan(t) = x/4 ; x = 4tan(t) cot(t) = 4/x ; x = 4/cot(t) thats all of them thatll use the x....

  40. amistre64
    • 5 years ago
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    you know how the trig functions are defined?

  41. anonymous
    • 5 years ago
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    Not really.

  42. anonymous
    • 5 years ago
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    What do I do for this problem though? \

  43. amistre64
    • 5 years ago
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    the names that are given to them arent all that important, but since they got em, might as well use em. sin is simply the ration of the opposite side from the angle divided by the hypotenuse: usually abbreviated as opp/hyp.

  44. amistre64
    • 5 years ago
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    you just express x or y in terms of the trig functions... as far as I can tell

  45. amistre64
    • 5 years ago
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    ...simply the *ratio* of the .....

  46. amistre64
    • 5 years ago
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    oonce you know sin and cos, the rest are just made with these two...

  47. anonymous
    • 5 years ago
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    How would I find x and y then?

  48. amistre64
    • 5 years ago
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    you would need more information to actually solve for x and y; you would need to know the measurement of an angle, or another side so that you could work things out. But what you have is not enough information to get any "solid" answers for the rest of it.

  49. anonymous
    • 5 years ago
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    That's all it shows in the problem.

  50. anonymous
    • 5 years ago
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    no angle or anything.

  51. amistre64
    • 5 years ago
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    then it cant expect you to "solve" for any values, just want you to be able to express what x and y are in terms of the rest of the information

  52. anonymous
    • 5 years ago
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    Here is another similar one if you can help with this one. Find the side labeled x. Assume a = 4.

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  53. anonymous
    • 5 years ago
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    How would I express them in terms then?

  54. amistre64
    • 5 years ago
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    this we have enough information to solve :)

  55. amistre64
    • 5 years ago
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    well, I wrote up all the ones I could think of for the x values up there.

  56. amistre64
    • 5 years ago
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    if a = 4 then sin(60) = x/4 sin(60) is one of those basic angles you learn in trig class and are expected to memorize. the sin(60) = sqrt(3)/2 sqrt(3) x ----- = --- 2 4 2sqrt(3) = x is as far as you can go without a calculator

  57. anonymous
    • 5 years ago
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    Oh.. ok I get this one. But the other has me confused :P

  58. anonymous
    • 5 years ago
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    It just says "Express x in terms of trigonometric ratios of θ."

  59. anonymous
    • 5 years ago
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    Same for y!

  60. amistre64
    • 5 years ago
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    the other one just says: (a) Express x in terms of trigonometric ratios of θ. (b) Express y in terms of trigonometric ratios of θ. it doesnt say "solve" for any particular value. Just use what you know about the trig functions to be able to express x or y in terms of them

  61. amistre64
    • 5 years ago
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    like this: sin(t) = x/y ; x = y * sin(t) csc(t) = y/x ; x = y/csc(t) tan(t) = x/4 ; x = 4tan(t) cot(t) = 4/x ; x = 4/cot(t)

  62. anonymous
    • 5 years ago
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    How would I go about doing that? I tried the X ones you put up, but they don't seem to work.

  63. amistre64
    • 5 years ago
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    do you have anymore options :)

  64. anonymous
    • 5 years ago
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    nope :(

  65. amistre64
    • 5 years ago
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    then I would just skip it and move to the next question..... :)

  66. amistre64
    • 5 years ago
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    lifes to short to worry about 1 trig problem that they dont give you enough information about right ?

  67. anonymous
    • 5 years ago
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    Haha... Thanks so much for your help. I'm actually learning math more from you than my stupid professor lol :)

  68. amistre64
    • 5 years ago
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    lol .... well, I do have the time.....alot of time :)

  69. anonymous
    • 5 years ago
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    Are you in college or what?

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