Question

Vectors question

Answer 1

For part A, I got theta = 84.6 degrees.
I'm a little confused on part B...

Answer 2

It wouldn't be this, would it?
\[(84.6\sqrt{113}, \sqrt{113})\]

Answer 3

Open study is so empty right now :O

Answer 4

Don't see a part A or B in the image

Answer 5

Angle B = 180° - 83° - 42° = 55°
(Angle sum property)

Answer 6

<B=180-(42+83)=180-125=55
use sine formula to find other sides.

Answer 7

\[\frac{ a }{ \sin A }=\frac{ b }{ \sin B}=\frac{ c }{ \sin C }\]

Answer 8

Sorry guys! I uploaded the wrong image. Here is the right one :P

Answer 9

I get 131.2 degrees

Answer 10

For part A

Answer 11

For part B the magnitude will be the square root of (49+64) or square root of 113

Answer 12

Dang :/
how'd you get 131.2?
Hey, at least I got part B right!!
Part II in part B... would that be (84.6113√113, √113)?

Answer 13

I mean (84.6(sqrt113), sqrt113)

Answer 14

part iib will be sqrt(113)cos(131.2),sqrt(113)sin(131.2)

Answer 15

As for 131.2 The dot product is -7

Answer 16

So -7=sqrt(113)*1*cos(theta)

Answer 17

cos^-1(-7/sqrt(113))=131.2

Answer 18

Isn't the dot product 1?
-7 + 8...

Answer 19

Oh! I see. That's where I messed up :P

Answer 20

Can you help with part C?

Answer 21

So it would look like...
-7 = sqrt113 * sqrt34 * costheta
And solve for theta? Did I set that up correctly?

Answer 22

Dot product is wrong but otherwise yes

Answer 23

Not -7 but -7*5+8*3

Answer 24

Okay, thanks! You've been extremely helpful, I appreciate it :)

Answer 25

No problem.

Answer 26

@Evoker
For part iib, should I multiply out the sqrt113?

Or just leave it like this: \[(\sqrt{113}\cos131.19, \sqrt{113}\sin131.19)\]sqrt(113)cos(131.2),sqrt(113)sin(131.2)