+0

+1
284
3
+580

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
+129799
+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

Jun 10, 2022
#2
+580
+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
+129799
+1

OK....good deal   !!!!

CPhill  Jun 10, 2022