Posted by Rachel on .
Two forces are pushing an ice shanty along the Ice. One has a magnitude of 330 lb in a direction due east. The other force has a magnitude of 110 lb in a direction 54 degrees east of north. What are the magnitude and direction of the resulant force?

Trig..please help! 
Reiny,
method 1: state them as vectors in component form
first : 330 due east
= (330cos 0, 330 sin 0) = (330, 0)
second( 110cos36°, 110sin36°) = (88.99 , 64.66)
add them to get (418.99, 64.66)
magnitude = √(418.99^2 + 64.66) = 423.95
direction: tanØ = 64.66/418.99
Ø = 8.77°
or a bearing of N 8.77° E
method 2:
construction the parallogram with the diagonal representing the magnitude
I get an obtuse angled triangle with sides 110 and 330 and the contained angle of 144° ...... (90+54)
by the cosine law:
r^2 = 110^2 + 330^2  2(110)(330)cos144
...
r = 423.95 , same as above
Use the Sine Law to find the direction of the resultant.