Question

two ships leave a port at the same time. the first ship sails on a bearing of 40 deg at 18 knots (nautical miles per hour) and the second ship on a bearing of 130 deg at 26 knots. how far apart are they after 1.5 hours?

Explanation:

Step1: Calculate the distance each ship travels

The first ship's speed $v_1 = 18$ knots and time $t = 1.5$ hours, so distance $d_1=v_1t = 18\times1.5=27$ nautical - miles. The second ship's speed $v_2 = 26$ knots and time $t = 1.5$ hours, so distance $d_2=v_2t=26\times1.5 = 39$ nautical - miles.

Step2: Find the angle between their paths

The bearing of the first ship is $40^{\circ}$ and the second is $130^{\circ}$, so the angle $\theta$ between their paths is $\theta=130 - 40=90^{\circ}$.

Step3: Use the Pythagorean theorem

Since the angle between their paths is $90^{\circ}$, if the distance between them is $D$, by the Pythagorean theorem $D=\sqrt{d_1^{2}+d_2^{2}}$. Substitute $d_1 = 27$ and $d_2 = 39$ into the formula: $D=\sqrt{27^{2}+39^{2}}=\sqrt{729 + 1521}=\sqrt{2250}=15\sqrt{10}\approx47.43$ nautical - miles.

Answer:

$15\sqrt{10}\approx47.43$ nautical - miles