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Mathematics 15 Online
OpenStudy (anonymous):

Find the angle between the given vectors to the nearest tenth of a degree. u = <8, 7>, v = <9, 7> @agent0smith

OpenStudy (amistre64):

\[|u|~|v|~cos~\alpha=u\cdot v\] solve for alpha

OpenStudy (agent0smith):

a · b = |a| × |b| × cos(θ) Find a*b first, like in the last question.

OpenStudy (anonymous):

so 8*9 and 7*7?

OpenStudy (amistre64):

yes, and add the results

OpenStudy (agent0smith):

8*9 + 7*7 = |a| × |b| × cos(θ) find |a| and |b|, just use pythagoras to find their magnitudes

OpenStudy (anonymous):

72+49=121

OpenStudy (anonymous):

so from there what would i do>

OpenStudy (agent0smith):

find |a| and |b|, just use pythagoras to find their magnitudes

OpenStudy (agent0smith):

121 = |a| × |b| × cos(θ) All you need to do is find the magnitude of a and b, then find cosθ

OpenStudy (agent0smith):

For a vector w = <a, b> \[\Large |w| = \sqrt{a^2+b^2}\]ie use pythag basically.

OpenStudy (anonymous):

i got like 87

OpenStudy (agent0smith):

For what?

OpenStudy (agent0smith):

u = <8, 7> Find |u| v = <9, 7> Find |v|

OpenStudy (anonymous):

i feel so dumb

OpenStudy (agent0smith):

haha why?

OpenStudy (anonymous):

i am so lost

OpenStudy (agent0smith):

If you had w = <4, 2> and need |w| \[\Large |w| = \sqrt{4^2+2^2} = \sqrt{20}\]

OpenStudy (agent0smith):

Find |u| and |v| in the same way.

OpenStudy (anonymous):

u= \[\sqrt{113}\] v=\[\sqrt{130}\]

OpenStudy (agent0smith):

Good, so \[\Large 121 = \sqrt{113} \sqrt{130} × \cos(θ)\]now find cos(θ)

OpenStudy (anonymous):

how lol

OpenStudy (agent0smith):

Could you do it if it was this?\[\Large 4 = 2 \times 3 \times \cos \theta\] If so, do the same thing!

OpenStudy (agent0smith):

Just get cosθ by itself.

OpenStudy (anonymous):

so would it be 3.4 or something close to that

OpenStudy (agent0smith):

No, show me or tell me what you did.

OpenStudy (anonymous):

i did sqrt of 113 timess sqrt of 130 then the answer i got i divided by 121

OpenStudy (agent0smith):

If you had 4=2×3×cosθ you would divide both sides by 6 right?

OpenStudy (anonymous):

yes

OpenStudy (agent0smith):

So with \[\Large 121 = \sqrt{113} \sqrt{130} × \cos(θ)\]do the same thing... you might notice the mistake you made.

OpenStudy (anonymous):

okay so when i i multiply the squre roots together i get 35.9, then i divided that by 121 and got 3.3 do i have to find the cos of that

OpenStudy (agent0smith):

If you had \[\Large 2 = 5 x\]you would divide both sides by 5 right? To get \[\Large \frac{ 2 }{ 5} = x\] You're doing things in the wrong order.

OpenStudy (agent0smith):

You're doing something weird with the square roots, actually... 35.9?? http://www.wolframalpha.com/input/?i=sqrt113+*+sqrt130

OpenStudy (anonymous):

121=121.2*cos(θ)

OpenStudy (agent0smith):

Yes, now find the angle theta.

OpenStudy (anonymous):

would i divide it now?

OpenStudy (agent0smith):

Yes, then find theta.

OpenStudy (anonymous):

i got when i divided 0.9983498

OpenStudy (agent0smith):

0.9983498 = cosθ Find θ

OpenStudy (agent0smith):

Use a calculator and the inverse cosine button

OpenStudy (anonymous):

Inverse cosine right!

OpenStudy (anonymous):

holy poop please tell me is that haha i've been here trying to figure it out

OpenStudy (anonymous):

i got 3.292 so rounded 3.3

OpenStudy (agent0smith):

Looks about right

OpenStudy (anonymous):

YAY finally lololol THANK YOU SO MUCH!! you are amazing

OpenStudy (agent0smith):

Haha you're welcome :)

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