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.
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!
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