## anonymous one year ago given the triangle below find sec x tan x and cos z in fraction form xz 25 yz 15

1. NotTim

I don't know the answer, but you should post the triangle for other people.

2. anonymous

|dw:1433534714159:dw|

3. anonymous

#nottim

4. Nnesha

@ <<---- to tag some @name

5. Nnesha

@ <<---- to tag some @ name

6. Nnesha

|dw:1433535491052:dw| it's right triangle right ????

7. Nnesha

$\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }$

8. Nnesha

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

ok

10. anonymous

@Nnesha

11. Nnesha

^ right so what did you got for 3rd side ?

12. anonymous

im really behind and i just need answers i Know it sound rude but im jugling 10 assignment

13. anonymous

at the same time

14. Nnesha

welcome to openstudy! that's not gonna happen you have to participate :-)

15. Nnesha

alright substitute a and c value solve for b :-)

16. Nnesha

$\huge\rm 15^2 + b^2 = 25^2$ b = ?

17. anonymous

90

18. anonymous

@Nnesha

19. Nnesha

nope how did you get 90 ?

20. Nnesha

take square of 25 and 15

21. anonymous

625 and 225

22. Nnesha

$\huge\rm a^2+ 225 = 625$ solve for a move 225 to the right side

23. anonymous

then subtract 225

24. Nnesha

yes right

25. Nnesha

then take square root

26. anonymous

20

27. Nnesha

ys right

28. Nnesha

|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

can we do another'

30. Nnesha

did you get that one ? cos = ? tan = ?

31. Nnesha

remember you need tan x and cos z

32. anonymous

i got it

33. Nnesha

alright

34. anonymous

|dw:1433539551579:dw|

35. anonymous

@Nnesha

36. Nnesha

sin^ theta =what ? remember the identity ?

37. Nnesha

sin^2 theta =what ? remember the identity ?

38. anonymous

its a different question

39. Nnesha

yes ik :-)

40. anonymous

|dw:1433539731456:dw|

41. anonymous

i have to simplify

42. Nnesha

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

no

44. anonymous

i realy need the answer i only have a couple hours to turn in everything

45. Nnesha

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

ok i understand

47. anonymous

this is a different one

48. anonymous

|dw:1433540518518:dw|

49. anonymous

@Nnesha

50. anonymous

ill be serious

51. Nnesha

alright what's the statement ?

52. anonymous

angle j

53. anonymous

?

54. Nnesha

is it right triangle ?

55. anonymous

no

56. anonymous

is it 35degrees? @Nnesha

57. Nnesha

did you apply law of cosines or law of sines ?

58. anonymous

cosine

59. anonymous

was it right?

60. Nnesha

gimme a sec i'll check

61. Nnesha

law of cosines when they gave you one angle check your question again

62. anonymous

then its 7

63. Nnesha

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

64. anonymous

thats the angle tho

65. Nnesha

what's the question seems like you mssing something look at the original question

66. Nnesha

are*

67. anonymous

what is the measure of angle j in the triangle

68. Nnesha

take a screenshot

69. Nnesha

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

|dw:1433541529143:dw|

71. Nnesha

yep now we can solve it thanks!

72. anonymous

sorry

73. Nnesha

nope 35 isn't right you have to use law of sines

74. Nnesha

$\huge\rm \frac{ \sin J}{ j } = \frac{ \sin I }{ i }$ plug in numbers solve for J

75. Nnesha

j is opposite side of angle J|dw:1433541845296:dw|

76. anonymous

$(\sin 10/10)$

77. Nnesha

sin is not 10 j side = 10

78. Nnesha

$\huge\rm \frac{ \sin J}{ j } = \frac{ \sin I }{ i }$ angle I = ? side i = ? j side = 10

79. anonymous

i = 17

80. Nnesha

yes right so plug in

81. Nnesha

and angle I = ?

82. anonymous

102

83. Nnesha

yes right

84. Nnesha

$\huge\rm \frac{ \sin J}{ 10 } = \frac{ \sin (102) }{ 17 }$ solve for angle J

85. anonymous

83

86. Nnesha

i didn't get 83

87. anonymous

oh sorry 97

88. Nnesha

i didn't get 97

89. anonymous

what did you get?

90. Nnesha

91. anonymous

im new to this it will take a long tim

92. anonymous

7

93. Nnesha

it's okay

94. Nnesha

how did you get 7 ? :o

95. Nnesha

did you use calculator ?

96. anonymous

yes

97. Nnesha

what did u put into the calculator ? parentheses are really important

98. anonymous

sin/ 10=sin(102)/17

99. Nnesha

sin J**

100. Nnesha

http://www.wolframalpha.com/ use this one