+0  
 
+1
291
3
avatar+5250 

Need help FAST!!!

 

 

Disregard ssm as it means the answer is in something I do not have.

rarinstraw1195  Sep 20, 2017
 #1
avatar+86859 
+2

Writing the vector components thusly :

 

A  = < 2cos 0° , 2sin 0° >

B  =  < 3.75 cos 90°, 3.75 sin 90° >

C  =  < 2.5 cos 180°. 2.5 sin 180° >

D = < 3 cos 270°, 3 sin 270° >

 

Sum the x components  = 2cos 0° + 3.75cos 90° + 2.5cos 180° + 3cos 270°  = 2 + 0 - 2.5 + 0  =  -.5

 

Sum the y components  = 2sin 0° + 3.75sin 90° +  2.5 sin 180° + 3 sin 270° = 0 + 3.75 + 0 - 3  = .75

 

Resultant vector  =  < x , y >  =  < -.5, .75 >     →  in the second quadrant

 

Resultant magnitude  = sqrt ( x^2 + y^2 )  =  sqrt [ (-.5)^2  + (.75)^2 ] ≈  . 90134 km

 

Direction  =   tan-1 (y/x)  = tan-1 ( .75 / -.5)  = tan-1 ( -3/2) = tan-1 (-1.5)  ≈ -56.31°

 

However....this is a second quadrant angle so, with respect to due west, it is 56.31° north of due west 

[i.e.  123.69°  in standard position  ]

 

 

 cool cool cool

CPhill  Sep 20, 2017
 #2
avatar+5250 
+1

Thank you CPhill!

rarinstraw1195  Sep 20, 2017
 #3
avatar+86859 
0

No prob, rarinstraw.....!!!!

 

 

 

cool cool cool

CPhill  Sep 20, 2017

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