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

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse