Two cruises leave the same port. Cruise A travels at 40km/h. Cruise B travels at 37km/h. The angle between their paths is 62 degrees. How far are the two cruises apart three hours later?

Answer 1

To solve this problem, we can use the law of cosines to find the distance between the two cruises.

Let's call the distance they are apart after 3 hours "d".

Using the law of cosines, we have:

d^2 = (40km)^2 + (37km)^2 - 2(40km)(37km)cos(62°)

Simplifying the equation, we have:

d^2 = 1600km^2 + 1369km^2 - 2(40km)(37km)cos(62°)

d^2 = 2969km^2 - 2(40km)(37km)cos(62°)

d^2 = 2969km^2 - 2(40km)(37km)(0.475)

d^2 = 2969km^2 - 888.2km^2

d^2 = 2080.8km^2

Taking the square root of both sides, we get:

d = √2080.8km^2

d ≈ 45.6 km

Therefore, the two cruises are approximately 45.6 km apart after 3 hours.