Posted by G0al on Tuesday, April 30, 2013 at 10:55pm.
I made a sketch
labeled end of first trip A, end 2nd trip B to get triangle OAB
OA=4 , AB = 18 and angle BAO = 64° by simple geometry
OA^2 = 18^2+4^2 - 2(4)(8)cos64 , by cosine law
..
OA = 16.6395 m
by sine law: Let angle AOB= Ø
sinØ/18 = sin64/16.6395
...
Ø = 76.5 or 180-76.5
Ø = 76.5 or 103.5
looking at my sketch, I chose 103.5°
so the new direction angle is 103.5+17= 120.5°
Maginitude is 16.6395 m and direction is 120.5°
or
u = (4cos17 , 4sin17)
3v = (18cos133, 18sin133)
u + 3v = (4cos17 , 4sin17) + (18cos133, 18sin133)
= (-8.45 , 14.338)
maginitude = √( (-8.45)^2 , 14.338^2 ) = 16.6395
direction angle
tanØ = 14.338/-8.45
Ø = 120.5°