By using a Scaled phasor Diagram determine the resultant force when the following currents are added. (I will attach a picture in the comment). What I need is from the method I have used clearly explain the method/technique used and to validate the solution/ method by using an alternative mathematical method.

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By using a Scaled phasor Diagram determine the resultant force when the following currents are added. (I will attach a picture in the comment). What I need is from the method I have used clearly explain the method/technique used and to validate the solution/ method by using an alternative mathematical method.

Mathematics
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here we can try to develop your formulas
please wait I'm computing your sum...

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here are my steps:
\[\begin{gathered} 20\cos \left( {60\pi t} \right)\cos \left( {\pi /4} \right) - 20\sin \left( {60\pi t} \right)\sin \left( {\pi /4} \right) + \hfill \\ \hfill \\ + 30\cos \left( {60\pi t} \right)\cos \left( {\pi /10} \right) - 30\sin \left( {60\pi t} \right)\sin \left( {\pi /10} \right) = \hfill \\ \hfill \\ = \cos \left( {60\pi t} \right)\left\{ {20\cos \left( {\pi /4} \right) + 30\cos \left( {\pi /10} \right)} \right\} - \hfill \\ \hfill \\ - \sin \left( {60\pi t} \right)\left\{ {20\sin \left( {\pi /4} \right) + 30\sin \left( {\pi /10} \right)} \right\} \hfill \\ \end{gathered} \]
now, we can write: \[\left\{ \begin{gathered} 20\cos \left( {\pi /4} \right) + 30\cos \left( {\pi /10} \right) = A\cos \alpha \hfill \\ 20\sin \left( {\pi /4} \right) + 30\sin \left( {\pi /10} \right) = A\sin \alpha \hfill \\ \end{gathered} \right.\]
using the subsequent values: \[\cos \left( {\pi /4} \right) = \sin \left( {\pi /4} \right) = \frac{1}{{\sqrt 2 }}\]
and:
\[\begin{gathered} \cos \left( {\pi /10} \right) = \frac{{\sqrt {10 + \sqrt {20} } }}{4}, \hfill \\ \hfill \\ \sin \left( {\pi /10} \right) = \frac{{\sqrt 5 - 1}}{4} \hfill \\ \end{gathered} \]
I rewrite those values:\[\begin{gathered} \cos \left( {\pi /10} \right) = \frac{{\sqrt {10 + \sqrt {20} } }}{4}, \hfill \\ \hfill \\ \sin \left( {\pi /10} \right) = \frac{{\sqrt 5 - 1}}{4} \hfill \\ \end{gathered} \]
\[\begin{gathered} \sin \left( {\pi /10} \right) = \frac{{\sqrt 5 - 1}}{4} \hfill \\ \cos \left( {\pi /10} \right) = \frac{{\sqrt {10 + \sqrt {20} } }}{4}, \hfill \\ \end{gathered} \]
\[\Large \begin{gathered} \sin \left( {\pi /10} \right) = \frac{{\sqrt 5 - 1}}{4} \hfill \\ \hfill \\ \cos \left( {\pi /10} \right) = \frac{{\sqrt {10 + \sqrt {20} } }}{4}, \hfill \\ \end{gathered} \]
I hope you see them correctly
we can rewrite those equations as below: \[\Large \left\{ \begin{gathered} \frac{{20}}{{\sqrt 2 }} + 30\frac{{\sqrt {10 + \sqrt {20} } }}{4} = A\cos \alpha \hfill \\ \hfill \\ \frac{{20}}{{\sqrt 2 }} + 30\frac{{\sqrt 5 - 1}}{4} = A\sin \alpha \hfill \\ \end{gathered} \right.\]
then we square both of those equations and we add them together, so we can write: \[\large {A^2} = {\left( {\frac{{20}}{{\sqrt 2 }} + 30\frac{{\sqrt {10 + \sqrt {20} } }}{4}} \right)^2} + {\left( {\frac{{20}}{{\sqrt 2 }} + 30\frac{{\sqrt 5 - 1}}{4}} \right)^2}\]
whereas divding side by side those equations, we find: \[\Large \tan \alpha = \frac{{A\sin \alpha }}{{A\cos \alpha }} = \frac{{\frac{{20}}{{\sqrt 2 }} + 30\frac{{\sqrt 5 - 1}}{4}}}{{\frac{{20}}{{\sqrt 2 }} + 30\frac{{\sqrt {10 + \sqrt {20} } }}{4}}}\]
@Michele_Laino thank you for your help. Sorry to be a pain but would you be able to hand write the first part for me and upload a picture just so i can try to understand some of the terminology of what you have put. e.g. hfill and \
no, I'm sorry I'm not able to hand write my answer, nevertheless I'm able to write a PDF file, and attach it using the "Attach FIle" button
from my preceding 2 formulas, you can find both the amplitude A, and the angle \alpha, so you can write your answer as follows: \[\Large {i_1} + {i_2} = A\cos \left( {60\pi t + \alpha } \right)\]
please wait, I'm doing that computation...
I got: \[\Large \begin{gathered} A = 48.67Amperes \hfill \\ \alpha = 0.5\;radians \hfill \\ \end{gathered} \]
so we can write: \[\Large {i_1} + {i_2} \cong 48.7\cos \left( {60\pi t + 0.5} \right)\]
please note that we have the subsequent equivalence: \[\Large 0.5\;{\text{radians}} = 28.75\;{\text{degrees}}\]
That would be helpful if you could put it as a PDF as im not use to how this is laid out. I normally have to type it up in word and it takes ages (see attached). That was why I suggested handwritten as it what i find easiest. But we all have our preferences.
@Michele_Laino you deserve more than 2 medals D:! All that Latex!
thanks! :) @UsukiDoll
Michele does
@carlj0nes ok! I start to write your PDF file, please wait, it will take some time
texmaker and then use Latex --> PDF I think it was F5 . I forgot.
just a note at the beginning should that of been 32 cos not 30 cos
That's right! I use TexMik with TeXnicCenter
ok! I update my answer with your value @carlj0nes
just a note at the beginning should that of been 32 cos not 30 cos
here is my PDF file:
what do you think about that file?
I saw "in order to that" @______@
is it correct? @UsukiDoll
doesn't sound right
why?
"In order to [do] that"
ok! thanks! I'm updating my file @UsukiDoll
A simple computation shows us...
[Therefore,] we can write
compute [the following] ratio
following the suggestions by @UsukiDoll I have rewritten my file, here is it:
[Then],
known angles [we have], [Next,] we get,
comma splice! period after Amplitude A A simple computation (no s) Again, after a simple computation, we get: delete So. add[ As a result, ]the requested sum of the two currents is
here my new updated version of the preceding PDF file, I have to be grateful to @UsukiDoll for her suggestions and comments to my english grammar!
I still see that comma splice
please what is a comma splice?
I'll take a screenshot and use paint to add my marks AHHAAHAHHAHAHA
ok!
comma splice is a comma between two complete sentences. It's the most aggravating error for English Professors to see on essays
ok! Please show where is the comma splice in my file
Thank you for all you help it is much appreciated. I am about to post another question and i would like to invite both of you.
I have to sleep early. I have to help shop blah

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