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?
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.