A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Please help me finish this precalculus 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!
anonymous
 one year ago
Please help me finish this precalculus 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\tan^{1} (\frac{ y_2y_1 }{ x_2x_1 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I use tan^1= y2y1/x2x1 to find a degree value for cos(theta)= 8/sqrt(365)? Isnt that the slope formula? @saseal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439085666228:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y2y1 / x2x1 gives you the Opposite/Adjacent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.08(4)/(32) = 4/1 =4 Okay @saseal

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0tan^1 (4) for the angle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for catching that @jim_thompson5910 . Is my denominator correct though?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, 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

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you apply the arccos function to both sides

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So would it be theta= arccos(19/(sqrt(365))? Would I just have to solve for that? @jim_thompson5910

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2now evaluate arccos(19/sqrt(365)) with a calculator

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2some calculators use \(\Large \cos^{1}\) in place of arccos

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 6.009 degrees, or 6 degrees if I round it to the nearest tenth. Thats one of my answer choices @jim_thompson5910

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank 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

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you're welcome

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[u=2 i4j\] \[v=3i8j\] \[u.v=\left( 2i4j \right)\left( 3i8j \right)=\left( 2 \right)\left( 3 \right)+\left( 4 \right)\left( 8 \right)\] =6+32=38 \[\left u \right=\sqrt{\left( 2 \right)^2+\left( 4 \right)^2}=\sqrt{20}=2\sqrt{5}\] \[\left v \right=\sqrt{\left( 3 \right)^2+\left( 8 \right)^2}=\sqrt{73}\] \[u.v=\left u \right\left v \right\cos \theta,~where ~\theta~is~the~\angle~\between~u~and~v\] \[38=2\sqrt{5}\sqrt{73}\cos \theta \] \[\cos \theta=\frac{ 19 }{ \sqrt{5}\sqrt{73} }\] \[\theta=\cos^{1} \left( \frac{ 19 }{ \sqrt{365} } \right)\]
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.