Two ships leave the same port. Ship A travels at a constant speed of 30 km/h. Ship B travels at a constant speed of 37 km/h. The angle between their paths 62 degrees. How far apart are the two ships from eachother three hours later?

The distance between them is 105.07 km

Step-by-step explanation:

First we need to find the distance travelled by each ship.

To find this distance, we should use the formula:

distance = speed * time

For ship A, we have:

distance1 = 30 * 3 = 90 km

For ship B, we have:

distance2 = 37 * 3 = 111 km

Now, to find the distance between the ships, we can use the law of cosines, where a, b and c are the respectively the distance travelled by ship A, the distance travelled by ship B and the distance between them:

c^2 = a^2 + b^2 - 2*a*b*cos(62)

c^2 = 90^2 + 111^2 - 2*90*111*0.4695

c^2 = 11040.39

c = 105.07 km

The distance between them is 105.07 km

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.