SOLUTION: Consider a vector for which a= 48º30' and whose magnitude is 32.6. What is the horizontal component? (Enter you answer rounded to one decimal place.) Thank you in advanced.

Algebra ->  Vectors -> SOLUTION: Consider a vector for which a= 48º30' and whose magnitude is 32.6. What is the horizontal component? (Enter you answer rounded to one decimal place.) Thank you in advanced.      Log On


   

Question 984221: Consider a vector for which a= 48º30' and whose magnitude is 32.6. What is the horizontal component? (Enter you answer rounded to one decimal place.)
Thank you in advanced.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
To find the horizontal component (x), use this formula x = r*cos(theta)

r = magnitude (ie the radius; or distance from the origin to the tip of the vector)
theta = angle

We're given r = 32.6 and theta = 48º30' = 48.5 degrees (note: 30 arcminutes= 30/60 = 0.5 degrees)

x = r*cos(theta)
x = 32.6*cos(48.5)
x = 21.601413571833
x = 21.6

The x component is approximately 21.6