Asked by Anonymous
                vector a is 8 ft., vector b is 2 ft, and theta is at 41 degrees. find the resultant and the direction.
I drew a diagram that shows a connecting with b (head to tail), and they make the 41 degrees. I use law of cosines and got about 6ft, which doesn't sound right to me.
my work:
r^2 = 8^2 + 2^2 - 2(8)(2)cos41
r^2 = 43.849
r = 6.621 ?
I was thinking the resultant would be larger than vector a. Also, how do you find the direction (angle) of the resultant?
Thanks!
            
        I drew a diagram that shows a connecting with b (head to tail), and they make the 41 degrees. I use law of cosines and got about 6ft, which doesn't sound right to me.
my work:
r^2 = 8^2 + 2^2 - 2(8)(2)cos41
r^2 = 43.849
r = 6.621 ?
I was thinking the resultant would be larger than vector a. Also, how do you find the direction (angle) of the resultant?
Thanks!
Answers
                    Answered by
            Reiny
            
    Usually the given angle is the angle between the two vectors, so after you place them end to end, I saw a triangle with sides 8 and 2 and a contained angle of 139º ( I placed the vector a horizontally)
Now by the cosine law,
R^2 = 64 + 4 - 2(8)(2)cos139
R = 9.6 (9.5995)
Let ß be the angle between the resultant R and the vector of length 8
sinß/2 = sin139/9.5995
ß = 7.856º degrees up from the horizontal.
    
Now by the cosine law,
R^2 = 64 + 4 - 2(8)(2)cos139
R = 9.6 (9.5995)
Let ß be the angle between the resultant R and the vector of length 8
sinß/2 = sin139/9.5995
ß = 7.856º degrees up from the horizontal.
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