Help solving this problem.

Problem

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)

Guest

Guest

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

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

3[b][e] = 349

A= 3[b][e] as given

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

A= 349

leesajohnson

Posts: 208
Joined: Thu Dec 31, 2015 7:11 am
Location: London
Reputation: -33

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!
Guest

Guest wrote:Help solving this problem.

Problem

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)

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

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

Desired vector C:
$$\vec{C}=\vec{A}+\vec{B}$$

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

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

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

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

I hope I have helped!

Baltuilhe

Posts: 29
Joined: Fri Dec 14, 2018 3:55 pm
Reputation: 21