Physics Help?

Question

A plane undergoes two consecutive displacements. The first is 67 km at 18.3 degrees west of north, and the second is 144 km at 59.7 degrees east of north. At what angle EAST OF NORTH is the plane's total displacement?

^ To find the displacement, I use law of cosine: squareroot(144^2 + 67^2 - 2*144*67*cos102) = 170.9879 km
Then I used law of sine to find the angle opposite to 67 km: ((sinX/(67)) = ((sin102)/(170.9879)) = 22.5369 degrees
Subtract it from 59.7 degrees and add 180 degrees and got 217.1631 degrees east of north as my answer.

I'm not sure if I'm doing this right. The angle seems too large for an answer. Can anyone tell me what I'm doing wrong?

Answer 1

You have done everything right except for the last part. The angle opposite 144 km is 102º: it is part of a triangle that contains the initial angle (18.3º) and the second angle (59.7º). Its value is then 180º-18.3º-59.7º=102º. Using the law of sines you got the angle opposite the 67km side as 22.537º This leaves the angle opposite the 144km leg = 180º-22.537º-102º=55.46º This last angle is made up of the initial angle of 18.3º west of north, and the final angle of x degrees east of north. Therefore your final result should be 55.46º - 18.3º = 37.16º