# a ship starts from port at a bearing of South 32 degrees East. It travels 8 miles in this direction before turning to a bearing of North 14 degrees East. It travels 15 miles in this direction, before finally turning to a bearing of South 72 degrees East and travels another 4 miles in this direction. How far is the ship from the port?

## I recommend doing this graphically.

## I haven tried doing it graphically using the law of sines but am still unable to come up with a consistent answer. It is also asking what is the bearing of the ship to the port from the final point?

## graphically is not the law of sines...

## ok thanks for the help?

## Given: AB = 8mi[148o]CW, BC = 15[14o], CD = 4[108o].

X = 8*sin148+15*sin14+4*sin108 = 11.7 miles.

Y = 8*cos148+15*cos14+4*cos108 = 6.5i miles.

D = 11.7 + 6.5i = 13.4mi[61] CW.

## D = 11.7 +6.5i = 13.4mi[61o] = AD.

DA = 13.4[61+180] = 13.4mi[241o] CW = bearing from final point to port.

## To find the distance of the ship from the port, we can break down the ship's journey into different segments and use trigonometry to calculate the distances.

1. Ship's first segment: It travels 8 miles on a bearing of South 32 degrees East.

To calculate the horizontal distance covered, we need to find the component of the distance traveled in the eastward direction.

Eastward distance = 8 * cos(32 degrees)

Eastward distance ≈ 6.78 miles (rounded to two decimal places)

To calculate the vertical distance covered, we need to find the component of the distance traveled in the southward direction.

Southward distance = 8 * sin(32 degrees)

Southward distance ≈ 4.23 miles (rounded to two decimal places)

2. Ship's second segment: It travels 15 miles on a bearing of North 14 degrees East.

We need to find the eastward and northward distances.

Eastward distance = 15 * cos(14 degrees)

Eastward distance ≈ 14.69 miles (rounded to two decimal places)

Northward distance = 15 * sin(14 degrees)

Northward distance ≈ 3.96 miles (rounded to two decimal places)

3. Ship's third segment: It travels 4 miles on a bearing of South 72 degrees East.

We need to find the eastward and southward distances.

Eastward distance = 4 * cos(72 degrees)

Eastward distance ≈ 1.30 miles (rounded to two decimal places)

Southward distance = 4 * sin(72 degrees)

Southward distance ≈ 3.86 miles (rounded to two decimal places)

Now, let's add up all the eastward and westward distances as well as the northward and southward distances:

Eastward distance = 6.78 + 14.69 + 1.30 ≈ 22.77 miles (rounded to two decimal places)

Northward distance = 3.96

Southward distance = 4.23 + 3.86 ≈ 8.09 miles (rounded to two decimal places)

To find the overall distance from the port, we can use the Pythagorean theorem:

Distance = √(Eastward distance^2 + (Northward distance - Southward distance)^2)

Distance = √(22.77^2 + (3.96 - 8.09)^2)

Distance ≈ √(519.43 + (-16.32)^2)

Distance ≈ √(519.43 + 266.13)

Distance ≈ √(785.56)

Distance ≈ 28 miles (rounded to two decimal places)

Therefore, the ship is approximately 28 miles away from the port.