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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

User Koperko
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1 Answer

5 votes

Answer:

37.59 nautical miles

Step-by-step explanation:

Distance = Speed x Time

The speed of the first ship = 12 knots

Thus, the distance covered after 1.5 hours


\begin{gathered} =12*1.5 \\ =18\text{ miles} \end{gathered}

The speed of the second ship = 22 knots

Thus, the distance covered after 1.5 hours


\begin{gathered} =22*1.5 \\ =33\text{ miles} \end{gathered}

The diagram representing the ship's path is drawn and attached below:

The angle at port = 90 degrees.

The triangle is a right triangle.

Using Pythagorean Theorem:


\begin{gathered} c^2=a^2+b^2 \\ c^2=18^2+33^2 \\ c^2=324+1089 \\ c^2=1413 \\ c=\sqrt[]{1413} \\ c=37.59\text{ miles} \end{gathered}

The two ships are 37.59 nautical miles apart after 1.5 hours.

two ships leave a port at the same time. the first ship sails on a bearing of 55° at-example-1
User Takis
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