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!

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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!

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\[\tan^{-1} (\frac{ y_2-y_1 }{ x_2-x_1 }\]
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
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imagine it this way
Okay. @saseal
y2-y1 / x2-x1 gives you the Opposite/Adjacent
-8-(-4)/(3-2) = -4/1 =-4 Okay @saseal
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)
tan^-1 (-4) for the angle
Thank you for catching that @jim_thompson5910 . Is my denominator correct though?
yes it is
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
you apply the arccos function to both sides
So would it be theta= arccos(19/(sqrt(365))? Would I just have to solve for that? @jim_thompson5910
now evaluate arccos(19/sqrt(365)) with a calculator
some calculators use \(\Large \cos^{-1}\) in place of arccos
I got 6.009 degrees, or 6 degrees if I round it to the nearest tenth. Thats one of my answer choices @jim_thompson5910
correct
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
you're welcome
\[u=2 i-4j\] \[v=3i-8j\] \[u.v=\left( 2i-4j \right)\left( 3i-8j \right)=\left( 2 \right)\left( 3 \right)+\left( -4 \right)\left( -8 \right)\] =6+32=38 \[\left| u \right|=\sqrt{\left( 2 \right)^2+\left( -4 \right)^2}=\sqrt{20}=2\sqrt{5}\] \[\left| v \right|=\sqrt{\left( 3 \right)^2+\left( -8 \right)^2}=\sqrt{73}\] \[u.v=\left| u \right|\left| v \right|\cos \theta,~where ~\theta~is~the~\angle~\between~u~and~v\] \[38=2\sqrt{5}\sqrt{73}\cos \theta \] \[\cos \theta=\frac{ 19 }{ \sqrt{5}\sqrt{73} }\] \[\theta=\cos^{-1} \left( \frac{ 19 }{ \sqrt{365} } \right)\]

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