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anonymous

  • one year ago

In right ∆ABC, angle B = 90°, angle C = 40°, and angle BC = 10. What are the other two side lengths of the triangle? A) AC = 13.1, AB = 8.4 B) AC = 18.7, AB = 15.8 C) AC = 14.3, AB = 10.2 D) AC = 15.6, AB = 11.9

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  1. anonymous
    • one year ago
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    Could really use the help, don't have to give me the answer outright just need to know how to solve this problem, can someone please guide me in the right direction? thank you.

  2. Nnesha
    • one year ago
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    |dw:1434457799463:dw| \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

  3. Nnesha
    • one year ago
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    so you need one angle and one side to find the other side of right triangle after that you can apply Pythagorean theorem

  4. anonymous
    • one year ago
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    @Nnesha Ahh so I solve it using SOHCAHTOA which is what my teacher taught me awhile back, I haven't done this sort of stuff in forever, might need some more assistance haha

  5. Nnesha
    • one year ago
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    yes alright!

  6. Nnesha
    • one year ago
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    as you can there is one angle and 2 sides in SOHCAHTOA formula i posted above so you need one angle and one side to solve for 3rd variable cos = adjacent over hyp so we to use cos function \[\huge\rm \cos (40) = \frac{ 10 }{ h }\] h = hypotenuse solve for h

  7. Nnesha
    • one year ago
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    |dw:1434458317996:dw| adjacent side that touch the angle

  8. anonymous
    • one year ago
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    @Nnesha I got D as my answer

  9. Nnesha
    • one year ago
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    how did you get D ?

  10. Nnesha
    • one year ago
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    \[\huge\rm \cos (40) = \frac{ 10 }{ h }\] solve for h and it's 40 degree

  11. Nnesha
    • one year ago
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    do you know how to solve for h ??

  12. anonymous
    • one year ago
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    @Nnesha Yea, I ended up getting 7.66

  13. Nnesha
    • one year ago
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    what is cos(40) make sure mode is 'n degree

  14. anonymous
    • one year ago
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    @Nnesha I got cos(40)=0.76604=10/h then I rearranged that, then multiplied 0.7660 x 10 and that was what I got, im trying to get an idea from this

  15. Nnesha
    • one year ago
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    why did multiply ??

  16. Nnesha
    • one year ago
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    you*

  17. anonymous
    • one year ago
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    @Nnesha https://www.mathsisfun.com/algebra/trig-finding-side-right-triangle.html

  18. Nnesha
    • one year ago
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    what about it ?

  19. Nnesha
    • one year ago
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    \[\huge\rm .766=\frac{ 10 }{ h }\] you should multiply both sides by H not by 10

  20. Nnesha
    • one year ago
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    \[\huge\rm .766 \times h = \frac{ 10 }{\cancel h} \times \cancel{h}\]

  21. anonymous
    • one year ago
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    @Nnesha Then divide both sides by .766? I ended up getting 13.054 or 13.1

  22. Nnesha
    • one year ago
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    yes right

  23. Nnesha
    • one year ago
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    right |dw:1434460279833:dw| now you can apply Pythagorean theorem to find 3rd side of right triangle \[\huge\rm a^2 +b^2 =c^2\] plug in values and solve for 3rd side remember c is the longest side of right triangle which is hypotenuse

  24. anonymous
    • one year ago
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    alright I did 10^2+b^2=13.1^2 then got 100+b^2=171.61 then subtracted 100 from both sides got b^2=71.61, I squared both sides and got B=8.4622

  25. Nnesha
    • one year ago
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    yes right

  26. Nnesha
    • one year ago
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    b^2 = 71.61 so b=what ?

  27. anonymous
    • one year ago
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    8.4622

  28. Nnesha
    • one year ago
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    right

  29. anonymous
    • one year ago
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    so its A, thank you so much haha, idk if you would be willing to help me with one more problem considering how long it took awhile to complete this haha

  30. Nnesha
    • one year ago
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    i'll try my best :-)

  31. anonymous
    • one year ago
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    @Nnesha haha ok thank you, I appreciate it In ∆ABC, angle B is a right angle and tan A = 13/16 . What is angle C ?

  32. Nnesha
    • one year ago
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    what's the definition of tan ?? look at my first comment tan equal what ?

  33. anonymous
    • one year ago
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    opposite over adjacent

  34. Nnesha
    • one year ago
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    yes now draw a right triangle

  35. anonymous
    • one year ago
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    @Nnesha Alright done

  36. Nnesha
    • one year ago
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    |dw:1434461030870:dw| which side is 13 and which one is 16 ?

  37. anonymous
    • one year ago
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    opposite should be 13 and adjacent should be 16?

  38. Nnesha
    • one year ago
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    yes right u should write 13 and 16 ?

  39. anonymous
    • one year ago
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    @Nnesha It looks funny to me, the adjacent side is longer than the opposite side?

  40. Nnesha
    • one year ago
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    that doesn't matter hypotenuse should be bigger than other 2 sides doesn't matter a is longer or b

  41. Nnesha
    • one year ago
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    which side is opposite of angle A ?

  42. anonymous
    • one year ago
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    @Nnesha the side that's 13

  43. Nnesha
    • one year ago
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    yes so |dw:1434461354773:dw|where u should write 13 ? and where suppose to be 16 ?

  44. anonymous
    • one year ago
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    between A and B would be the 13 and between B and C would be the 16

  45. Nnesha
    • one year ago
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    |dw:1434461430223:dw| yep right now tanC = opposite over adjacent substitute values solve for C

  46. anonymous
    • one year ago
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    @Nnesha Im having difficulty understanding what to do >_< would I multiply both sides by one of the numbers or?

  47. Nnesha
    • one year ago
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    did u substitute values of opposite and adjacent side let me know u get after that

  48. anonymous
    • one year ago
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    Ok yea I just push in Tan(C)=13/16

  49. Nnesha
    • one year ago
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    well wait we did it wrong

  50. Nnesha
    • one year ago
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    tan A = 13/16 not tan C

  51. Nnesha
    • one year ago
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    so it should be like this |dw:1434462085204:dw|

  52. anonymous
    • one year ago
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    @Nnesha Ahhh ok, I knew something was off, so then Tan(C)=16/13

  53. Nnesha
    • one year ago
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    now tan = opposite over adjacent \[\huge\rm tan (C)= \frac{ 16 }{ 13 }\] yes that's right

  54. Nnesha
    • one year ago
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    so u need to find invs of tan \[\huge\rm C = \tan^{-1} \frac{ 16 }{ 13 }\] use calculator

  55. anonymous
    • one year ago
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    Alright, I got 0.8884

  56. Nnesha
    • one year ago
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    make sure mode 'n degree

  57. anonymous
    • one year ago
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    @Nnesha Yea, it keeps giving me the same answer for some reason C=0.88847979

  58. Nnesha
    • one year ago
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    bec mode of calculator is on rad so it should be on degree

  59. Nnesha
    • one year ago
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    hit rad and then type that

  60. Nnesha
    • one year ago
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    http://prntscr.com/7hm54j

  61. anonymous
    • one year ago
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    I got this now C=50.906141113771

  62. anonymous
    • one year ago
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    @Nnesha

  63. Nnesha
    • one year ago
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    looks good

  64. anonymous
    • one year ago
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    @Nnesha Thank you for everything, I really appreciate it

  65. Nnesha
    • one year ago
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    my pleasure

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