A ship travels due north for 25 miles then changes course to a bearing of 300°. It travels on this path for 29 miles. How far is the ship from where it started?

Sum the east (x-) and north (y-) components.

x=∑D cos(θ)
=25*cos(90°)+29 cos(90-300°)
=0+29cos(-210)
=-(29/2)√3

y=∑D sin(θ)
=25*sin(90°)+29sin(90-300°)
=25+29sin(-210°)
=25+29sin(30°)
=25+14.5
=39.5

Distance
=sqrt(x^2+y^2)
=...

To find out how far the ship is from where it started, we can use the concept of vectors and vector addition. Let's break down the ship's movement into two separate vectors.

Vector A represents the ship's initial motion due north. It has a magnitude (length) of 25 miles and a direction of 0° (north is conventionally considered as 0°).

Vector B represents the ship's change in course. It has a magnitude of 29 miles and a direction of 300°.

Now, we need to add these two vectors together to find the resultant vector, which represents the total displacement of the ship.

To add two vectors, we can break them down into their horizontal (x) and vertical (y) components and then add the corresponding components together.

For Vector A:
- The x-component is 0 (since it only moves vertically).
- The y-component is 25 miles (since it moves directly north).

For Vector B:
- The x-component is 29 * cos(300°) (to get the horizontal component).
- The y-component is 29 * sin(300°) (to get the vertical component).

Using trigonometry, we find that:
- cos(300°) = 0.866 (approximately)
- sin(300°) = -0.5 (approximately)

So, the x-component of Vector B is 29 * 0.866 = 25.034 miles (approximately), and the y-component is 29 * -0.5 = -14.5 miles (approximately).

Now, we can add the x-components and y-components separately.
- The x-component is 0 + 25.034 = 25.034 miles.
- The y-component is 25 + (-14.5) = 10.5 miles.

Using the Pythagorean theorem, we can find the magnitude of the resultant vector (the distance from where the ship started) which is the square root of the sum of the squares of the x and y components:
- Magnitude = sqrt((25.034^2) + (10.5^2)) = sqrt(626.144 + 110.25) = sqrt(736.394) = 27.115 miles (approximately).

Therefore, the ship is approximately 27.115 miles from where it started.

To find the distance of the ship from where it started, we can use the concept of vector addition.

First, let's visualize the ship's path.

The ship initially travels due north for 25 miles. This can be represented as a vector pointing straight up.

Then, the ship changes course to a bearing of 300° and travels along this path for 29 miles. This can be represented as a vector pointing 300° clockwise from due north.

Now, let's use trigonometry to find the horizontal and vertical components of this vector.

The horizontal component can be found by multiplying the magnitude of the vector (29 miles) by the cosine of the angle (300°).

Horizontal component = 29 miles * cos(300°)

The vertical component can be found by multiplying the magnitude of the vector (29 miles) by the sine of the angle (300°).

Vertical component = 29 miles * sin(300°)

Now, let's calculate these values.

Horizontal component = 29 miles * cos(300°) = 29 miles * 0.866 = 25.094 miles (rounded to three decimal places)

Vertical component = 29 miles * sin(300°) = 29 miles * (-0.5) = -14.5 miles

Since the ship initially traveled due north for 25 miles, the vertical component of its position is 25 miles.

To find the total displacement of the ship, we add the horizontal and vertical components of the vector:

Total displacement = √((Horizontal component)^2 + (Vertical component)^2)

Total displacement = √((25.094 miles)^2 + (-14.5 miles)^2)

Total displacement = √(630.47 miles^2 + 210.25 miles^2)

Total displacement = √(840.72 miles^2)

Total displacement = 29 miles (rounded to the nearest mile)

Therefore, the ship is approximately 29 miles from where it started.