Two cruises leave the same port. Cruise A travels at 30 km/h. Cruise B travels at 37 km/h. The angle between their paths is 62 degrees . How far are the two cruises apart three hours later?
To find the distance between the two cruises, we can use the law of cosines. Let's call the distance between Cruise A and Cruise B after 3 hours as $d$. The distance traveled by Cruise A after 3 hours is $30 \, \text{km/h} \times 3 \, \text{h} = 90$ km. Similarly, the distance traveled by Cruise B after 3 hours is $37 \, \text{km/h} \times 3 \, \text{h} = 111$ km.
The law of cosines states that $c^2 = a^2 + b^2 - 2ab \cos(C)$, where $c$ is the length of the side opposite to angle $C$. We want to find $c$, so let's plug in the given values:
$c^2 = 90^2 + 111^2 - 2 \cdot 90 \cdot 111 \cdot \cos(62^\circ)$.
Using a calculator, we find that $\cos(62^\circ) \approx 0.436$.
$c^2 = 8100 + 12321 - 2 \cdot 90 \cdot 111 \cdot 0.436$.
$c^2 = 8042.92$.
Taking the square root of both sides, we find $c \approx 89.81$.
Therefore, the two cruises are approximately 89.81 km apart after 3 hours.