A ship sails 15.0 mi on a course S40°10’W and then 21.0 mi on a course N28°20’W. Find the distance and direction of the last position from the first.

convert to (x,y) values and add:

15@S40°10'W = (-9.675,-11.463)
21@N28°20'W = (-9.967,18.484)
sum = (-19.642,7.021)
or,
[email protected]°W

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how did you get the [email protected]°W?

To find the distance and direction of the last position from the first, we can use vector addition.

First, we need to convert the given courses into Cartesian coordinates (north is positive and east is positive):

Course S40°10'W can be converted as follows:
-40°10' from the west is equivalent to 180° - 40°10' = 139°50' from the east
Since it's south, the angle becomes negative, so -139°50'
The Cartesian coordinates are (15.0 * cos(-139°50'), 15.0 * sin(-139°50'))

Course N28°20'W can be converted as follows:
-28°20' from the west is equivalent to 180° - 28°20' = 151°40' from the east
The Cartesian coordinates are (21.0 * cos(-151°40'), 21.0 * sin(-151°40'))

Next, we can add the Cartesian coordinates together to find the total displacement:
(displacement_x, displacement_y) = ((15.0 * cos(-139°50')) + (21.0 * cos(-151°40')), (15.0 * sin(-139°50')) + (21.0 * sin(-151°40')))

The distance between the initial and final positions is given by the Pythagorean theorem:
distance = sqrt(displacement_x^2 + displacement_y^2)

Finally, the direction of the last position from the first can be found using the inverse tangent function:
direction = atan2(displacement_y, displacement_x)

By calculating these equations, you can determine the distance and direction of the last position from the first.