A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Please help me finish this pre-calculus question!! Find the angle between the given vectors to the nearest tenth of a degree. u = <2, -4>, v = <3, -8> This is how far I got: cos(theta)= u * v/ IIuII IIvII cos(theta)= <2, -4> * <3, -8>/ II<2, -4>II II<3, -8>II cos(theta)= (2)(-4) + (3)(-8)/ sqrt((2)^2 + (-4)^2)) sqrt((3)^2 + (-8)^2)) cos(theta)= -8+24/sqrt(20) sqrt(73) cos(theta)= 16/ sqrt(20) sqrt(73) cos(theta)= 16/2*sqrt(365) cos(theta)= 8/sqrt(365) I dont know how to get a degree value from this? Please help!

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\tan^{-1} (\frac{ y_2-y_1 }{ x_2-x_1 }\]

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I use tan^-1= y2-y1/x2-x1 to find a degree value for cos(theta)= 8/sqrt(365)? Isnt that the slope formula? @saseal

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1439085666228:dw|

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    imagine it this way

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay. @saseal

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    y2-y1 / x2-x1 gives you the Opposite/Adjacent

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    -8-(-4)/(3-2) = -4/1 =-4 Okay @saseal

  8. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you made a mistake here `cos(theta)= (2)(-4) + (3)(-8)/ sqrt((2)^2 + (-4)^2)) sqrt((3)^2 + (-8)^2))` the numerator should be (2)*(3) + (-4)*(-8)

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    tan^-1 (-4) for the angle

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you for catching that @jim_thompson5910 . Is my denominator correct though?

  11. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yes it is

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay, so if I simplify it further, my end result should be cos(theta)=19/sqrt(365). But I am stuck here because the question is asking for the angle between the given vectors. Im not sure what to do next to get that angle. @jim_thompson5910

  13. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you apply the arccos function to both sides

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So would it be theta= arccos(19/(sqrt(365))? Would I just have to solve for that? @jim_thompson5910

  15. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    now evaluate arccos(19/sqrt(365)) with a calculator

  16. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    some calculators use \(\Large \cos^{-1}\) in place of arccos

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got 6.009 degrees, or 6 degrees if I round it to the nearest tenth. Thats one of my answer choices @jim_thompson5910

  18. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    correct

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you so much for taking the time to help me with this question, its been nagging at me for quite a while now! Thanks a lot @jim_thompson5910

  20. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you're welcome

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[u=2 i-4j\] \[v=3i-8j\] \[u.v=\left( 2i-4j \right)\left( 3i-8j \right)=\left( 2 \right)\left( 3 \right)+\left( -4 \right)\left( -8 \right)\] =6+32=38 \[\left| u \right|=\sqrt{\left( 2 \right)^2+\left( -4 \right)^2}=\sqrt{20}=2\sqrt{5}\] \[\left| v \right|=\sqrt{\left( 3 \right)^2+\left( -8 \right)^2}=\sqrt{73}\] \[u.v=\left| u \right|\left| v \right|\cos \theta,~where ~\theta~is~the~\angle~\between~u~and~v\] \[38=2\sqrt{5}\sqrt{73}\cos \theta \] \[\cos \theta=\frac{ 19 }{ \sqrt{5}\sqrt{73} }\] \[\theta=\cos^{-1} \left( \frac{ 19 }{ \sqrt{365} } \right)\]

  22. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.