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?

Answer 1

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.