Answer:
Explanation:
From the information given:
Let assume that the drone first delivery went to a certain place U, 7.10miles from the warehouse in the S 25.0° W direction.
However, the drone thus proceeds to the second delivery place V which is 11.2 miles in the S 38.0° W direction.
Then, location U can be determined from a graphical point of view as follows:
In the negative x-direction from the warehouse


In the negative y-direction from the warehouse



Also: the position of V with respect to U can be determined as follows:
in the positive x-direction from the warehouse


In the negative y-direction from the warehouse:


Similarly, we will need to determine the position of V with respect to the warehouse.
i.e.
In the positive x-direction the warehouse

-6.44 + 8.83 ( since
is in the negative direction)
2.39 miles
In the positive y-direction the warehouse

( since both
are in the negative direction)
-9.89 miles
Therefore, from above, the distance emanating how far is the drone back to the starting point is :



D = 10.18 miles
The direction of the drone can be deduced by taking the tangent of the trigonometry;
i.e.




in the north_west direction of V