Adog searching for a bone walks 3.50 m south, then 8.20 m at an angle of 30.0° north of east, and finally 15.0 m west. use graphical techniques to find the dog’s resultant displacement vector.

First draw both axes then draw a vector of (theoretical) length 3.5 along the negative x axes. 
Then draw a "8.2" vector from that endpoint all the way up, 60 degrees from that point (90-30) This is because you make a 30 degree angle with the current vertical and then you will see that the angle in the triangle we are making is 90 degrees minus that 30 so we come up ith 60 degrees. 
You should make the 8.20 vector higher that the orginal 3.5 m vector starting point.
Next we will finally draw a 15 m vector from the endpoint of the 8.2 m vector that goes above the original one and to the left. This one should be drawn really far. 

We will solve for the sides. The first triangle we make is 30-60-90, so we must solve fot the vertical side which is just:
8.2/2 = 4.1 m 
Now solve for the opposite side using 4.1(sqrt(3)) = 7.1 m. So now we have enough to solve for the other triangle that the 15m line makes. 
the top side is 15 - 7.1 = 7.9 m 
while the right side is 4.1 - 3.5 = 0.6 m 
 The resultant displacement is the last side: 
√(7.9^2) + (0.6^2) = 7.9 m (ANSWER) 
To find the direction simply use: 
arcTan(y/x) = arcTan(0.6/7.9) = 4.3 degrees north of west (ANSWER 2)

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