SOLUTION: a ship sails due east from post A to post B for 6.9 miles. Then it turns in the direction N 42 degrees and 30' W and sails to port C which is 5.2 miles away from B. Find the distan

Algebra ->  Triangles -> SOLUTION: a ship sails due east from post A to post B for 6.9 miles. Then it turns in the direction N 42 degrees and 30' W and sails to port C which is 5.2 miles away from B. Find the distan      Log On


   

Question 1030578: a ship sails due east from post A to post B for 6.9 miles. Then it turns in the direction N 42 degrees and 30' W and sails to port C which is 5.2 miles away from B. Find the distance and bearing of the ship from A.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
a ship sails due east from post A to post B for 6.9 miles. Then it turns in the direction N 42 degrees and 30' W and sails to port C which is 5.2 miles away from B. Find the distance and bearing of the ship from A.
-------
Start at (0,0)
1st leg DATA: x-coordinate:: 6.9*cos(0) ; y-coordinate:: 6.9*sin(0)
------
2nd leg DATA: x-coordinate:: 5.2cos(132.5) ; y-coordinate:: 5.2sin(132.5)
-------
Final position::
x-coordinate = sum of x-coordinates = 3.39
y-coordinate = sum of y-coordinates = -3.51
----
Ans:
distance from A = sqrt[3.39^2 + 3.51^2] = 4.88 miles
bearing from A:: arctan(-3.51/3.39) = 90+46 = 136 degrees
--------------------
Cheers,
Stan H.