Two ships leave the same port at the same time. One ship sails on a course of 110 degree at 32 mi/h. The other sails on a course of 230 degree at 40 mi/h. Find the distance between them after 2 hours. Express your answer to the nearest mile.

To find the distance between the two ships after 2 hours, we can use the Law of Cosines.

Let's label the distance between the two ships as 'd'.

First, we need to find the angle between the two ships. The angle between the two ships is the sum of the angles made by each ship with the north.

For the first ship, sailing at a course of 110 degrees, the angle it makes with the north is 90 degrees - 110 degrees = -20 degrees (measured counterclockwise from the positive x-axis).

For the second ship, sailing at a course of 230 degrees, the angle it makes with the north is 90 degrees + 230 degrees = 320 degrees (measured counterclockwise from the positive x-axis).

Now, we can apply the Law of Cosines:

d² = a² + b² - 2abcos(C)

where a = 32 mi/h (speed of the first ship), b = 40 mi/h (speed of the second ship), and C = 320 degrees - (-20 degrees) = 340 degrees.

Plugging in the values into the formula, we get:

d² = (32)² + (40)² - 2(32)(40)cos(340 degrees)

simplifying further:

d² = 1024 + 1600 - 2560cos(340 degrees)

d² = 2624 - 2560cos(340 degrees)

Using a calculator, compute cos(340 degrees):

cos(340 degrees) ≈ 0.9397

Substitute the value of cos(340 degrees) into the equation:

d² = 2624 - 2560(0.9397)

d² ≈ 2624 - 2402.72

d² ≈ 221.28

Finally, take the square root of both sides to find the distance between the two ships after 2 hours:

d ≈ √(221.28)

d ≈ 14.88

Rounding to the nearest mile, the distance between the two ships after 2 hours is approximately 15 miles.

To find the distance between the two ships after 2 hours, we can use the law of cosines. The law of cosines states that, for a triangle with sides a, b, and c and angle C opposite side c:

c^2 = a^2 + b^2 - 2ab * cos(C)

In this case, we can consider the two ships and the distance between them as the sides of a triangle. Let's label the distance between them as side c, the speed of the first ship as a, and the speed of the second ship as b.

Since the two ships are leaving the same port at the same time, they will have sailed for the same duration of 2 hours. Therefore, the distance traveled by the first ship is 32 mi/h * 2 h = 64 miles, and the distance traveled by the second ship is 40 mi/h * 2 h = 80 miles.

We now have:
a = 64 miles
b = 80 miles

Next, let's find the angle C. The angle between the two ships can be thought of as the vector difference between their courses, which will be 230 degrees - 110 degrees = 120 degrees.

We have:
C = 120 degrees

Now we can apply the law of cosines to find the distance between the two ships:

c^2 = a^2 + b^2 - 2ab * cos(C)

Plugging in the values:
c^2 = 64^2 + 80^2 - 2 * 64 * 80 * cos(120 degrees)

Using a calculator to evaluate the above expression, we find:
c^2 = 4096 + 6400 + 8192 * (-0.5)
c^2 = 10496 - 8192 * 0.5
c^2 = 10496 - 4096
c^2 = 6400

Taking the square root of both sides, we find:
c = sqrt(6400)

Approximating to the nearest mile, we have:
c ≈ 80 miles

So, the distance between the two ships after 2 hours is approximately 80 miles.

or, since the angle between their travel is 120° use the law of cosines to find the 3rd side of the triangle:

d^2 = 64^2 + 80^2 - 2*64*80 cos 120°

both start at origin

Ship one
110 from north is 20 degrees below +x axis
x1 = 2 * 32 * cos 20
y1 = -2 * 32 * sin 20
Ship 2
230 from north is 40 deg below -x axis
x2 = -2 * 40 cos 40
y2 = -2 * 40 sin 40

d = sqrt [(x2-x1)^2 + (y2-y1)^2 ]