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avatar+578 

If the polar coordinates for A, B, and C are (4, pi/4), (3, 5pi/12) and (2, pi/12)

 

Calculate: AB^2, AC^2, and BC^2

 

I got: -32-32sqrt(3)i, -32+32sqrt(3)i, -16

 

I just want to check if anyone got the same

 Jun 10, 2022
edited by Vinculum  Jun 10, 2022
 #1
avatar+124525 
+2

A = ( 4cos[ pi/4 ]   , 4 sin [ pi /4 ])     =  (4cos 45° , 4sin 45°)  =      (4 / sqrt( 2)  , 4/sqrt ( 2 )  )

 

B =  (3 cos [ 5pi/12 ],  3sin [5pi/12 ] )  =  (3cos 75° , 3sin 75°)  = 

(  3 [  sqrt (3) - 1 ] / [ sqrt (8)]  ,  3 [ sqrt (3) + 1] / [sqrt (8)]   )

 

C = ( 2, pi/12)  =  (2 cos 15° , 2sin 15°) =   (2 sin 75° , 2cos 75°) =  

( 2 [sqrt (3) + 1]/ [sqrt (8)]  ,  2[sqrt (3) -1 ]  / {sqrt (8)  ] 

 

 

Vinculum ...   Do you want   AB^2  , AC^2  and BC^2  ??

 

Or do you mean  (AB)^2 , (AC)^2  and  (BC)^2   ??

 

Also....these are real points......there shouldn't be any  "i"    involved

 

 

cool cool cool

 Jun 10, 2022
 #2
avatar+578 
+1

Hey CPhill I was looking for AB^2, AC^2 and BC^2

 

Thank you very much for the answer, I really appreciate it. 

 

I think I can finish it from here!!!

Vinculum  Jun 10, 2022
 #3
avatar+124525 
+1

OK....good deal   !!!!

 

 

cool cool cool

CPhill  Jun 10, 2022

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