anonymous
  • anonymous
Find the angle between the given vectors to the nearest tenth of a degree. u = <2, -4>, v = <3, -8>
Trigonometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zzr0ck3r
  • zzr0ck3r
\[a\cdot b=|a|*|b|*cos(\theta), where\space \theta\space is\space the\space angle\space between\space the\space vectors\]
anonymous
  • anonymous
I tried everything, but I'm not getting the answer I need...
anonymous
  • anonymous
\[ \cos \theta = \frac {a \cdot b} {|a| |b| } \] take the arccos to find \[ \theta\]

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anonymous
  • anonymous
I know that's how you solve it, however my issue is that I am not getting the right answer. Obviously, I'm doing something wrong.
anonymous
  • anonymous
http://www.wikihow.com/Find-the-Angle-Between-Two-Vectors
zzr0ck3r
  • zzr0ck3r
\[a \cdot b = 38\\ |a|*|b| = \sqrt{4+16}*\sqrt{73}=2\sqrt{5}\sqrt{73 }\\\theta=\cos^{-1}(\frac{19}{\sqrt{5}\sqrt{73}})\approx0.10487\]
zzr0ck3r
  • zzr0ck3r
@Loveiskey18 do you understand?
anonymous
  • anonymous
\[ \Large \cos \theta = \frac {a \cdot b} {|a| |b| }\\ \Large \cos \theta = \frac {(2*3)+(-4*-8)} {\sqrt{2^2+3^2}*\sqrt{4^4+8^2} }\\ \]
zzr0ck3r
  • zzr0ck3r
your vectors are wrong on the bottom @cinar shuold be \[\sqrt{2^2+4^2}\\and\\\sqrt{3^2+8^2}\]
anonymous
  • anonymous
@zzr0ck3r Sorry I took so long, I was attempting in understanding this but I am not. I just keep getting stuck at the part where you input the vectors.
anonymous
  • anonymous
you are right
zzr0ck3r
  • zzr0ck3r
where I input the vectors? do you know what the dot product is, and the TWO different ways it is defined?
zzr0ck3r
  • zzr0ck3r
you need to know this before you can answer this.
zzr0ck3r
  • zzr0ck3r
@Loveiskey18 this will go allot faster if you answer:)
anonymous
  • anonymous
I'm sorry, I was attempting it on my own...again
zzr0ck3r
  • zzr0ck3r
ok just stay with me and ill walk you through it?
anonymous
  • anonymous
My ending results: arccos (38/√(13)√(-80))
zzr0ck3r
  • zzr0ck3r
nope
zzr0ck3r
  • zzr0ck3r
can you tell me what \[<2,-4>\cdot<3,-8>=?\]
anonymous
  • anonymous
arccos (38/√(-1040)
zzr0ck3r
  • zzr0ck3r
square root of negative number is the first thing that I can see is wrong
zzr0ck3r
  • zzr0ck3r
please just stick with me and stop posting answers:)
anonymous
  • anonymous
Laughing out loud, okay.
zzr0ck3r
  • zzr0ck3r
\[<2,-4>\cdot<3,-8>=?\]
zzr0ck3r
  • zzr0ck3r
I think you did this part already, its 38 right?
anonymous
  • anonymous
Yes.
zzr0ck3r
  • zzr0ck3r
do you understand \[\theta=cos^{-1}(\frac{a\cdot b}{|a||b|})\]
anonymous
  • anonymous
Yes.
zzr0ck3r
  • zzr0ck3r
so we need |a|*|b| correct?
anonymous
  • anonymous
Correct.
zzr0ck3r
  • zzr0ck3r
\[a=<2,-4> \\|a| =\sqrt{2^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}\sqrt{5}=2\sqrt{5}\\correct?\]
anonymous
  • anonymous
Yes.
zzr0ck3r
  • zzr0ck3r
\[b=<3,-8>\\|b|=\sqrt{3^2+(-8)^2}=\sqrt{9+64}=\sqrt{73}\]correct?
anonymous
  • anonymous
Yes.
zzr0ck3r
  • zzr0ck3r
if there is anything you dont understand just tell me, I am in no rush
anonymous
  • anonymous
I acknowledge everything completely.
zzr0ck3r
  • zzr0ck3r
ok then we have\[\theta=\cos^{-1}(\frac{a\cdot b}{|a|*|b|})=cos^{-1}(\frac{38}{2\sqrt{5}\sqrt{73}})=\cos^{-1}(\frac{19}{\sqrt{5*73}})=\cos^{-1}(\frac{19}{\sqrt{365}})\]
zzr0ck3r
  • zzr0ck3r
\[\approx0.10487\]
anonymous
  • anonymous
To be honest I do understand all of this, and actually I already obtained that answer before I asked for help, however I wasn't sure. The problem is... that is not the answer I am looking for, well it is. I just do not know how to convert it into degrees or the appropriate answer: My choices: 3.0° 6.0° -7.0° 16.0° Nonetheless, I am very much grateful for you putting in the time and energy to help me.
zzr0ck3r
  • zzr0ck3r
we got radians as the answer\[radian\space answer*\frac{180}{\pi}=degree\space answer\]
zzr0ck3r
  • zzr0ck3r
\[.10487*\frac{180}{\pi}\approx6.0081\approx6\]
zzr0ck3r
  • zzr0ck3r
next time you are confused about radians to degree imagine if you have pi, then you know that is 180 degrees well what can we multiply pi by to get 180? pi(180/pi) = 180
zzr0ck3r
  • zzr0ck3r
and if you have 180 degrees you know that is pi well what can we multiply 180 by to get pi? 180*(pi/180)=pi
zzr0ck3r
  • zzr0ck3r
ok so, after all that....are we good to go now?
anonymous
  • anonymous
Thank You so much :D Do you mind verifying a couple questions for me, please. Just tell me if they are right or wrong.

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