Radian Measurement

Radian Measurement

Postby Guest » Fri Feb 06, 2015 9:47 am

Help solving this problem.

Problem

Add the following vectors:


A = 3[b][e], θA = 1[c]4° and B = 2[c][d], θB = 3[a]8°

(Note: if PIN = 34659, then 3[b][e] = 349)

Round your answers to one decimal place.
Guest
 

Re: Radian Measurement

Postby Guest » Fri Feb 06, 2015 4:41 pm

Any comment about my problem?
Guest
 

Re: Radian Measurement

Postby Guest » Wed Feb 11, 2015 11:40 am

It is quite difficult to understand your problem. If it was clear to me I would help you.
Guest
 

Re: Radian Measurement

Postby Guest » Sat Sep 12, 2015 1:55 pm

Your instructor should have assigned you a pin number. You have to have it to sold the problem.
Guest
 

Re: Radian Measurement

Postby leesajohnson » Mon Jun 27, 2016 4:25 am

3[b][e] = 349

A= 3[b][e] as given

so 3[b][e] = 349 = A

A= 349

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Re: Radian Measurement

Postby Guest » Wed Apr 10, 2019 5:58 pm

Sorry but I have no idea what "vector A= 3[b][e]" even means! Could you please explain the notation?

Nor do I understand how leesajohnson got "A= 349" when we wer told that A is vector not a number!
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Re: Radian Measurement

Postby Baltuilhe » Wed Apr 17, 2019 5:31 pm

Guest wrote:Help solving this problem.

Problem

Add the following vectors:


A = 3[ b][e], θA = 1[c]4° and B = 2[c][d], θB = 3[a]8°

(Note: if PIN = 34659, then 3[ b][e] = 349)

Round your answers to one decimal place.

Good afternoon!
PIN = 34659, then:
[a]=3
[ b]=4
[c]=6
[d]=5
[e]=9

So:
Data:
[tex]\begin{cases}\|\vec{A}\|=349&\theta_A=164^{\circ}\\\|\vec{B}\|=265&\theta_B=338^{\circ}\end{cases}[/tex]

Desired vector C:
[tex]\vec{C}=\vec{A}+\vec{B}[/tex]

Coordinates of vector A:
[tex]\|\vec{A_x}\|=340\times\cos\;164^{\circ}=-326,83\\
\|\vec{A_y}\|=340\times\sin\;164^{\circ}=93,72[/tex]

Coordinates of vector B:
[tex]\|\vec{B_x}\|=265\times\cos\;338^{\circ}=245,70\\
\|\vec{B_y}\|=265\times\sin\;338^{\circ}=-99,27[/tex]

Sum of vectors A and B:
[tex]\|\vec{C_x}\|=\|\vec{A_x}\|+\|\vec{B_x}\|=-81,13\\
\|\vec{C_y}\|=\|\vec{A_y}\|+\|\vec{B_x}\|=-5,55[/tex]

Now the solution:
[tex]\begin{cases}\|\vec{C}\|=\sqrt{81,13^2+5,55^2}=81,32\\
\theta_C=-\arctan\left(\dfrac{5,55}{81,13}\right)=-176,08^{\circ}\end{cases}[/tex]

I hope I have helped!

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