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anonymous

  • one year ago

given the triangle below find sec x tan x and cos z in fraction form xz 25 yz 15

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  1. NotTim
    • one year ago
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    I don't know the answer, but you should post the triangle for other people.

  2. anonymous
    • one year ago
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    |dw:1433534714159:dw|

  3. anonymous
    • one year ago
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    #nottim

  4. Nnesha
    • one year ago
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    @ <<---- to tag some @name

  5. Nnesha
    • one year ago
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    @ <<---- to tag some @ name

  6. Nnesha
    • one year ago
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    |dw:1433535491052:dw| it's right triangle right ????

  7. Nnesha
    • one year ago
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    \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

  8. Nnesha
    • one year ago
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    yes it's a right triangle find 3rd side of right triangle Pythagorean theorem \[\huge\rm a^2 + b^2 = c^2\] c is the longest side of right triangle which is hypotenuse solve for 3rd side

  9. anonymous
    • one year ago
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    ok

  10. anonymous
    • one year ago
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    @Nnesha

  11. Nnesha
    • one year ago
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    ^ right so what did you got for 3rd side ?

  12. anonymous
    • one year ago
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    im really behind and i just need answers i Know it sound rude but im jugling 10 assignment

  13. anonymous
    • one year ago
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    at the same time

  14. Nnesha
    • one year ago
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    welcome to openstudy! that's not gonna happen you have to participate :-)

  15. Nnesha
    • one year ago
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    alright substitute a and c value solve for b :-)

  16. Nnesha
    • one year ago
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    \[\huge\rm 15^2 + b^2 = 25^2\] b = ?

  17. anonymous
    • one year ago
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    90

  18. anonymous
    • one year ago
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    @Nnesha

  19. Nnesha
    • one year ago
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    nope how did you get 90 ?

  20. Nnesha
    • one year ago
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    take square of 25 and 15

  21. anonymous
    • one year ago
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    625 and 225

  22. Nnesha
    • one year ago
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    \[\huge\rm a^2+ 225 = 625\] solve for a move 225 to the right side

  23. anonymous
    • one year ago
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    then subtract 225

  24. Nnesha
    • one year ago
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    yes right

  25. Nnesha
    • one year ago
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    then take square root

  26. anonymous
    • one year ago
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    20

  27. Nnesha
    • one year ago
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    ys right

  28. Nnesha
    • one year ago
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    |dw:1433537872991:dw| now use this information \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\] cos = adjacent over hyp tan = opposite over adjacent remember sec = 1/cos

  29. anonymous
    • one year ago
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    can we do another'

  30. Nnesha
    • one year ago
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    did you get that one ? cos = ? tan = ?

  31. Nnesha
    • one year ago
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    remember you need tan x and cos z

  32. anonymous
    • one year ago
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    i got it

  33. Nnesha
    • one year ago
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    alright

  34. anonymous
    • one year ago
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    |dw:1433539551579:dw|

  35. anonymous
    • one year ago
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    @Nnesha

  36. Nnesha
    • one year ago
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    sin^ theta =what ? remember the identity ?

  37. Nnesha
    • one year ago
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    sin^2 theta =what ? remember the identity ?

  38. anonymous
    • one year ago
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    its a different question

  39. Nnesha
    • one year ago
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    yes ik :-)

  40. anonymous
    • one year ago
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    |dw:1433539731456:dw|

  41. anonymous
    • one year ago
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    i have to simplify

  42. Nnesha
    • one year ago
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    ik i'm asking do you know the sin cos identities most important one is \[\huge\rm sin^2 \theta + \cos^2 =1\] solve it for sin^2

  43. anonymous
    • one year ago
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    no

  44. anonymous
    • one year ago
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    i realy need the answer i only have a couple hours to turn in everything

  45. Nnesha
    • one year ago
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    sin^2 + cos ^2 = 1 solve for sin^2 please first of all i don't like giving direct answers (enabling ) 2nd i'm not allowed to give you direct answer http://openstudy.com/code-of-conduct

  46. anonymous
    • one year ago
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    ok i understand

  47. anonymous
    • one year ago
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    this is a different one

  48. anonymous
    • one year ago
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    |dw:1433540518518:dw|

  49. anonymous
    • one year ago
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    @Nnesha

  50. anonymous
    • one year ago
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    ill be serious

  51. Nnesha
    • one year ago
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    alright what's the statement ?

  52. anonymous
    • one year ago
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    angle j

  53. anonymous
    • one year ago
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    ?

  54. Nnesha
    • one year ago
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    is it right triangle ?

  55. anonymous
    • one year ago
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    no

  56. anonymous
    • one year ago
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    is it 35degrees? @Nnesha

  57. Nnesha
    • one year ago
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    did you apply law of cosines or law of sines ?

  58. anonymous
    • one year ago
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    cosine

  59. anonymous
    • one year ago
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    was it right?

  60. Nnesha
    • one year ago
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    gimme a sec i'll check

  61. Nnesha
    • one year ago
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    law of cosines when they gave you one angle check your question again

  62. anonymous
    • one year ago
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    then its 7

  63. Nnesha
    • one year ago
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    i din't say 35 is wrong .... for law of cosine you should have all 3 sides or at least one angle

  64. anonymous
    • one year ago
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    thats the angle tho

  65. Nnesha
    • one year ago
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    what's the question seems like you mssing something look at the original question

  66. Nnesha
    • one year ago
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    are*

  67. anonymous
    • one year ago
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    what is the measure of angle j in the triangle

  68. Nnesha
    • one year ago
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    take a screenshot

  69. Nnesha
    • one year ago
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    i'm saying they should gave u all 3 sides or 2 sides one angle that's not info to solve the question

  70. anonymous
    • one year ago
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    |dw:1433541529143:dw|

  71. Nnesha
    • one year ago
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    yep now we can solve it thanks!

  72. anonymous
    • one year ago
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    sorry

  73. Nnesha
    • one year ago
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    nope 35 isn't right you have to use law of sines

  74. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \sin J}{ j } = \frac{ \sin I }{ i }\] plug in numbers solve for J

  75. Nnesha
    • one year ago
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    j is opposite side of angle J|dw:1433541845296:dw|

  76. anonymous
    • one year ago
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    \[(\sin 10/10)\]

  77. Nnesha
    • one year ago
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    sin is not 10 j side = 10

  78. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \sin J}{ j } = \frac{ \sin I }{ i }\] angle I = ? side i = ? j side = 10

  79. anonymous
    • one year ago
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    i = 17

  80. Nnesha
    • one year ago
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    yes right so plug in

  81. Nnesha
    • one year ago
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    and angle I = ?

  82. anonymous
    • one year ago
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    102

  83. Nnesha
    • one year ago
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    yes right

  84. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \sin J}{ 10 } = \frac{ \sin (102) }{ 17 }\] solve for angle J

  85. anonymous
    • one year ago
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    83

  86. Nnesha
    • one year ago
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    i didn't get 83

  87. anonymous
    • one year ago
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    oh sorry 97

  88. Nnesha
    • one year ago
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    i didn't get 97

  89. anonymous
    • one year ago
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    what did you get?

  90. Nnesha
    • one year ago
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    how did you solve ? show your work please :(

  91. anonymous
    • one year ago
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    im new to this it will take a long tim

  92. anonymous
    • one year ago
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    7

  93. Nnesha
    • one year ago
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    it's okay

  94. Nnesha
    • one year ago
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    how did you get 7 ? :o

  95. Nnesha
    • one year ago
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    did you use calculator ?

  96. anonymous
    • one year ago
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    yes

  97. Nnesha
    • one year ago
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    what did u put into the calculator ? parentheses are really important

  98. anonymous
    • one year ago
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    sin/ 10=sin(102)/17

  99. Nnesha
    • one year ago
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    sin J**

  100. Nnesha
    • one year ago
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    http://www.wolframalpha.com/ use this one

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