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.
Add the two vectors.
To find the distance and direction of the last position from the first, we can use vector addition.
Step 1: Convert the given courses into Cartesian coordinates.
The first course, S40°10’W, can be converted to Cartesian coordinates (x1, y1) as follows:
x1 = 15.0 * sin(40°10’W)
y1 = -15.0 * cos(40°10’W)
The second course, N28°20’W, can be converted to Cartesian coordinates (x2, y2) as follows:
x2 = 21.0 * sin(28°20’W)
y2 = 21.0 * cos(28°20’W)
Step 2: Add the Cartesian coordinates together to find the resultant coordinates.
x = x1 + x2
y = y1 + y2
Step 3: Calculate the distance between the two positions using the Pythagorean theorem:
distance = sqrt(x^2 + y^2)
Step 4: Calculate the direction of the last position from the first using the inverse tangent function:
direction = arctan(y / x)
Step 5: Convert the direction from radians to degrees.
direction_degrees = direction * (180 / π)
Now let's plug in the values and calculate the answers:
x1 = 15.0 * sin(40°10’W) ≈ -9.64 mi
y1 = -15.0 * cos(40°10’W) ≈ -11.44 mi
x2 = 21.0 * sin(28°20’W) ≈ 9.39 mi
y2 = 21.0 * cos(28°20’W) ≈ -18.84 mi
x = x1 + x2 ≈ -9.64 mi + 9.39 mi ≈ -0.25 mi
y = y1 + y2 ≈ -11.44 mi + (-18.84 mi) ≈ -30.28 mi
distance = sqrt(x^2 + y^2) ≈ sqrt((-0.25 mi)^2 + (-30.28 mi)^2) ≈ 30.29 mi
direction = arctan(y / x) ≈ arctan((-30.28 mi) / (-0.25 mi)) ≈ 1.17 radians
direction_degrees = 1.17 * (180 / π) ≈ 67.06°
Therefore, the distance between the last position and the first position is approximately 30.29 miles, and the direction of the last position from the first position is approximately 67.06°.