## 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!

1. anonymous

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

2. anonymous

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

|dw:1439085666228:dw|

4. anonymous

imagine it this way

5. anonymous

Okay. @saseal

6. anonymous

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

7. anonymous

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

8. jim_thompson5910

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

tan^-1 (-4) for the angle

10. anonymous

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

11. jim_thompson5910

yes it is

12. anonymous

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

you apply the arccos function to both sides

14. anonymous

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

15. jim_thompson5910

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

16. jim_thompson5910

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

17. anonymous

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

correct

19. anonymous

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

you're welcome

21. anonymous

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