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April 18, 2014

April 18, 2014

Posted by **Rachel** on Sunday, January 6, 2013 at 10:18pm.

- Trig..please help! -
**Reiny**, Sunday, January 6, 2013 at 11:34pmmethod 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.

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