Two boats leave the same port at the same time. One travels at a speed of 31 mi/h in the direction N 50° E and the other travels at a speed of 20 mi/h in a direction S 70° E (see the figure). How far apart are the two boats after one hour? (Round your answer to the nearest mile.)

convert to rectangular form and subtract one from the other.

convert back to polar to get the distance.

Solving

To solve this problem, we can use the Law of Cosines. The Law of Cosines states that for any triangle with sides a, b, and c, and angle C opposite side c, the following equation holds:

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

In this case, we have two sides and the included angle for each boat's triangle. Let's calculate the distances traveled by each boat after one hour:

Boat A:
Distance = Speed * Time = 31 mph * 1 hour = 31 miles

Boat B:
Distance = Speed * Time = 20 mph * 1 hour = 20 miles

Now, we can calculate the angle C for each boat's triangle using the given directions:

For Boat A:
Angle C = 90° - 50° = 40°

For Boat B:
Angle C = 90° + 70° = 160°

Substituting the values of the distances and angles into the Law of Cosines:

For Boat A:
c^2 = (31)^2 + (20)^2 - 2(31)(20) * cos(40°)

For Boat B:
c^2 = (31)^2 + (20)^2 - 2(31)(20) * cos(160°)

Calculating each equation, we find:

For Boat A:
c^2 ≈ 961 + 400 - 1240 * cos(40°)

For Boat B:
c^2 ≈ 961 + 400 - 1240 * cos(160°)

Now, let's calculate each value:

For Boat A:
c^2 ≈ 961 + 400 - 992.97 ≈ 1368.03

For Boat B:
c^2 ≈ 961 + 400 - (-413.23) ≈ 1774.23

To find the distance between the two boats after one hour, we can use the concept of vector addition.

First, let's represent the two boats' velocities as vectors. The velocity of the first boat can be represented as V₁ = 31 mi/h at an angle of 50° east of north. The velocity of the second boat can be represented as V₂ = 20 mi/h at an angle of 70° east of south.

Next, we need to decompose these velocities into their north and east components. We can use trigonometry to find these components.

For the first boat, the north component (V₁n) can be found by multiplying the velocity magnitude (31 mi/h) by the cosine of the angle (50°):

V₁n = 31 mi/h * cos(50°)

For the first boat, the east component (V₁e) can be found by multiplying the velocity magnitude (31 mi/h) by the sine of the angle (50°):

V₁e = 31 mi/h * sin(50°)

Similarly, for the second boat, the north component (V₂n) can be found by multiplying the velocity magnitude (20 mi/h) by the cosine of the angle (20°):

V₂n = 20 mi/h * cos(20°)

For the second boat, the east component (V₂e) can be found by multiplying the velocity magnitude (20 mi/h) by the sine of the angle (20°):

V₂e = 20 mi/h * sin(20°)

Now, we can find the total north and east displacements for each boat by multiplying their respective north and east components by the time of one hour.

For the first boat, the north displacement (d₁n) can be found by multiplying the north component (V₁n) by 1 hour:

d₁n = V₁n * 1 hour

Similarly, for the first boat, the east displacement (d₁e) can be found by multiplying the east component (V₁e) by 1 hour:

d₁e = V₁e * 1 hour

For the second boat, the north displacement (d₂n) can be found by multiplying the north component (V₂n) by 1 hour:

d₂n = V₂n * 1 hour

Similarly, for the second boat, the east displacement (d₂e) can be found by multiplying the east component (V₂e) by 1 hour:

d₂e = V₂e * 1 hour

Finally, to find the distance between the two boats after one hour, we can use the Pythagorean theorem:

distance = √((d₂n - d₁n)² + (d₂e - d₁e)²)

By substituting the values we calculated, we can find the distance between the two boats after one hour.