Question

Two ships leave a harbor at the same time. One ship travels on a bearing S11°W at 14 miles per hour. The other ship travels on a bearing N75°E at 9 miles per hour. How far apart will the ships be after 3 ​hours?

Having a bit of trouble with this can someone please help

Answer 1

When you draw a diagram, you have a triangle

with sides of 30 mi (15 x 2) and 18 mi (9 x 2),

and an included angle of 116 degrees.

We are solving for the side opposite the 116 degree angle.

Now use the law of cosines:

d^2 = 30^2 + 18^2 - 2*30*18*cos 116

= 1124 + 473.44

= 1697.44 mi^2. So

d = 41.2 miles

Answer 2

Answer:

Distance=59.05mi

Step-by-step explanation:

The very first thing we need to do a drawing that will represent the situation (see picture attached, not drawn in scale).

From the drawing, we can see the relation between the angles and the path the ships are following. As you may see, the paths of the ships will build a triangle, which we will need to solve in order to find the distance between the two ships.

The first ship, the one traveling at 14mph, has traveled a distance of 42mi. We find this distance by multiplying the speed with the time:

Distance=speed*time

Distance= (14mph)(3h)=42mi

We can do the same with the second ship, the one traveling at 9 miles per hour.

Distance=(9mph)(3h)=27mi

Once we got those distances, we can now find the angle there is between the paths.

In the drawing, you can see that there is an angle of  105 degrees between the red line and the south axis and an additional 11° between the south axis and the blue line, giving us a total of 105°+11°=116°

With this information we can go ahead and use the cosine rule to find the distance between the two ships after 3 hours, this is the length of the green line.

$A^{2}=B^{2}+C^{2}-2BCcos($θ)

We can substitute the data in the drawing so we get:

$d^{2}=(27)^{2}+(42)^{2}-2(27)(42)cos(116^{o})$

which yields:

$d^{2}=3487.23$

when taking the square root to both sides we get:

$d=\sqrt{3487.23}=59.05mi$