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Question
Two ships leave a port at the same time. One goes 24 km/hr in the direction N 38° E and other travels 32 km/hr in the direction S 52° E. Find the distance between the ships at the end of 3 hrs.
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Solution
\[\text{ After three hours, let the ships be at P and Q respectively } . \]
\[\text{ Then }, \]
\[OP = 24 \times 3 = 72 km \text{ and } OQ = 32 \times 3 = 96 km\]
\[\text{ From figure, we have }\]
\[\angle POQ = 180° - \angle NOP - \angle SOQ\]
\[ = 180° - 38° - 52° \]
\[ = 90° \]
\[\text{ Now }, \]
\[\text{ Since OPQ is a right angled triangle }\]
\[ \therefore P Q^2 = O P^2 + O Q^2 \]
\[ \Rightarrow P Q^2 = {72}^2 + {96}^2 \]
\[ \Rightarrow P Q^2 = 5184 + 9216\]
\[ \Rightarrow P Q^2 = 14400\]
\[ \Rightarrow PQ = \sqrt{14400} = 120 km\]
\[\text{Hence, the distance between the ships after 3 hours is 120 km} .\]
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