Vectors help, can someone please help? easy 10 points

005 (part 1 of 2)
Tammy leaves the office, drives 33 km due
north, then turns onto a second highway and
continues in a direction of 32
◦
north of east
for 89 km.
What is her total displacement from the
office?
Answer in units of km

006 (part 2 of 2)
At what angle is her displacement? (Consider
east to be 0
◦
and north 90
◦
.)
Answer in units of

General method:
After choosing your axes and drawing a diagram illustrating the problem, divide any distance not given strictly along one of those axes into its components (which do lie along those axes), and add everything up so you have an x-term and a y-term, corresponding to the total x-displacement and the total y-displacement, respectively. These components will form a right triangle with the net displacement as the hypotenuse; you can then use a^2 + b^2 = c^2 to solve for the net displacement.

This question:
leg 1:
33km y-direction

leg 2:
89km 32◦ N of E (above the horizontal)
= 89km(cos32◦) x-direction + 89km(sin32◦) y-direction
= 75.5km x-direction + 47.2km y-direction

total y-displacement = 33km + 47.2km = 80.2km
total x-displacement = 75.5km
a^2 + b^2 = c^2
(80.2km)^2 + (75.5km)^2 = c^2
6432km^2 + 5700km^2 = c^2
12132km^2 = c^2
110.1km = c

Angle:
tanA = opposite side / adjacent side
tanA = 80.2km/75.5km
tanA = 1.06
A = tan^-1(1.06)
A = 46.7◦ North of East

OK, the first thing you need to learn about vectors math is how to reduce all the vectors into their XY components. Then you do the math on the components, not the vectors, to get two answers: X and Y. And after doing the math you reconstruct a vector as the answer using the Pythagoras Theorem on the X and Y answers. I'll do 005 to show how.

N = 33 km at theta = 90 deg (north) and E = 89 km at theta = 32 deg (ENE)

In vectors D = N + E where D is the displacement you're looking for.

XY

Nx = N cos(theta) = 33*0 = 0 and Ex = E cos(32) = 89*cos(radians(32)) = 75.47628056 = 75.5 km
Ny = 33*sin(90) = 33 and Ey = 89*sin(radians(32)) = 47.16281452 = 47.2 km

So X = 0 + 75.5 = 75.5 km
Y = 33 + 47.2 = 80.2 km

And from Pythagorus

D = sqrt(X^2 + Y^2) = sqrt(75.5^2 + 80.2^2) = 110.1466749 = 110.1 km at rho = ATAN(Y/X) = ATAN(80.2/75.5) = 46.73246213 = 46.7 deg. ANS.

You do the same steps for all the vector adds and subtracts. Note, as a vector, displacement has both magnitude and direction.