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(100angle30)*(-10angle-165)

Guest Jun 13, 2015

Best Answer 

 #7
avatar+889 
+10

Hi Melody

 

As Alan has said $$r\angle \theta$$ is a shorthand notation used mainly by electrical engineers.

It's actually shorthand for the complex number $$r(\cos \theta + \imath \sin \theta).$$

Written out in full, we would have

$$r_{1}\angle \theta_{1}\times r_{2}\angle\theta_{2}\\=r_{1}r_{2}(\cos\theta_{1}+\imath\sin\theta_{1})(\cos\theta_{2}+\imath\sin\theta_{2})\\=r_{1}r_{2}(\cos\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2}+\imath(\sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2}))\\=r_{1}r_{2}(\cos(\theta_{1}+\theta_{2})+\imath\sin(\theta_{1}+\theta_{2}))\\=r_{1}r_{2}\angle(\theta_{1}+\theta_{2}).$$

Bertie  Jun 15, 2015
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8+0 Answers

 #1
avatar+26328 
+10

Multiply the magnitudes together and add the angles:

 

100∠30*(-10∠-165) = -(100*10)∠(30-165) = -1000∠-135

.

Alan  Jun 14, 2015
 #2
avatar+91001 
0

Thanks Alan,

What kind of question would you use that for ?

Melody  Jun 15, 2015
 #3
avatar+26328 
+5

Electrical engineers commonly use this sort of representation for the magnitude and phase of ac electrical currents; but in general you could use it for any two-dimensional vectors expressed in polar coordinates.

.

Alan  Jun 15, 2015
 #4
avatar+91001 
0

Mmm, thanks Alan, I still don't have any concept of what it means.

Does it have a pictorial representation or a situation word problem that I could understand where it would be useful?

Melody  Jun 15, 2015
 #5
avatar+26328 
+5

If a vector is given in polar coordinates by (r, θ) then it can also be written as r∠θ (which electrical engineers do a lot!).

 

Multiplying two together is just a combined stretching (r1*r2) and rotating (θ1 + θ2) transformation.

.

Alan  Jun 15, 2015
 #6
avatar+91001 
0

okay, thanks Alan  

Melody  Jun 15, 2015
 #7
avatar+889 
+10
Best Answer

Hi Melody

 

As Alan has said $$r\angle \theta$$ is a shorthand notation used mainly by electrical engineers.

It's actually shorthand for the complex number $$r(\cos \theta + \imath \sin \theta).$$

Written out in full, we would have

$$r_{1}\angle \theta_{1}\times r_{2}\angle\theta_{2}\\=r_{1}r_{2}(\cos\theta_{1}+\imath\sin\theta_{1})(\cos\theta_{2}+\imath\sin\theta_{2})\\=r_{1}r_{2}(\cos\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2}+\imath(\sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2}))\\=r_{1}r_{2}(\cos(\theta_{1}+\theta_{2})+\imath\sin(\theta_{1}+\theta_{2}))\\=r_{1}r_{2}\angle(\theta_{1}+\theta_{2}).$$

Bertie  Jun 15, 2015
 #8
avatar+91001 
0

That makes good sense, Thanks Bertie  

Melody  Jun 15, 2015

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