SOLUTION: Consider a vector for which a= 48�30' and whose magnitude is 32.6. What is the horizontal component? (Enter you answer rounded to one decimal place.) Thank you in advanced.

Question 984221: Consider a vector for which a= 48�30' 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) (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 = 48�30' = 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
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