Question

two ships leave a port at the same time. the first ship sails on a bearing of 55° at 12 knots (natural miles per hour) and the second on a bearing of 145° at 22 knots. how far apart are they after 1.5 hours (round to the nearest nautical mile)

Answer 1

Given that one of the ships travels at 12 nautical miles per hour, then after 1.5 hours, it will travel 12*1.5 = 18 miles

The other ship will travel 22*1.5 = 33 miles

The angle of 145° is measured with respect to the positive x-axis. Then, respect the negative x-axis, its measure is 180° - 145° = 35°

We have to use trigonometric functions to find x1, y1, x2, and y2, as follows:

sin(55°) = y1/18

sin(55°) *18 = y1

14.7 = y1

cos(55°) = x1/18

cos(55°)*18 = x1

10.3 = x1

sin(35°) = y2/33

sin(35°)*33 = y2

18.9 = y2

cos(35°) = x2/33

cos(35°)*33 = x

-27 = x2 (the minus sign comes from the graph)

The distance between two points (x1, y1) and (x2, y2) is computed as follows:

`d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}`

Substituting with the values found,

`\begin{gathered} d=\sqrt[]{(-27-10.3)^2+(18.9-14.7)^2} \\ d=\sqrt[]{(-37.3)^2+4.2^2} \\ d=\sqrt[]{1391.29+17.64} \\ d\approx38\text{ miles} \end{gathered}`