PHYSICS 1 ..
posted by JOAN .
as she pick up her riders a bus driver traverses four successive displacements represented by the expression (6.30 b)i(4.00 b cos 40)i(4.00 sin 40)j+(3.00 b cos 50)i(3.00 b sin 50)j(5.00 b)j here b represents one city block,a convenient unit of distance of uniform size; i is east; and j is north. the displacements at 40 degree and 50 degree represent travel on roadways in the city that are at these angles to the main eastwest and northsouth streets.(a)draw a map of the successive displacements(b)what total distance did she travel? (c)compute the magnitude and direction of her total displacements.

add up all your x (east) distances. I assume that they are all in b units and
(4.00 sin 40)j
is a typo:
 6.3  4(.766) + 3(.643) = 7.44 b east
(or 7.44 b locks west)
Add up your y (north) distances:
 4(.643)  3(.766)  5 = 9.87 b North
(or 9.87 blocks south)
Total distance
6.3+4+4+3+3+5 = 25.3
Total displacement
7.44 west and 9.87 south
magnitude^2 = 7.44^2+ 9.87^2
so
magnitude = 12.4 b
direction
tan angle below x axis = 9.87/7.44
angle = 53 degrees below x axis
or
West 53 degrees South
or
217 degrees true on a compass 
The distance is 18.3. 6.3+4+3+5

Total distance is 6.3+4+3+5= 18.3 blocks
Magnitude is 12.4b
tan^1(9.9/7.4)= 53 but since it is in third quadrant, add 180 degrees to get a final angle of 233 degrees.
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