As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30°…
CBSE Class 10 Maths PYQ · Applications of Trig · Double Triangle · 5 Marks · March 2023 · Standard
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385 Marks · March 2023 · Standard
As observed from the top of a $75$ m high lighthouse from the sea-level, the angles of depression of two ships are $30^\circ$ and $60^\circ$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. (Use $\sqrt{3}= 1.73$)
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1 for correct figure PQ = Height of Light house = $75$ m $\angle XQS = \angle QSP = 30^\circ$ $\angle XQR = \angle QRP = 60^\circ$ R and S are position of ships. In $\triangle PQR$, $\frac{75}{PR} = \tan 60^\circ = \sqrt{3} \Rightarrow PR = \frac{75}{\sqrt{3}} = 25\sqrt{3}$ In $\triangle PQS$, $\frac{75}{PS} = \tan 30^\circ$ PS = $75\sqrt{3}$ $\therefore$ Distance between the ships, RS = PS – PR $= 75 \sqrt{3}-25 \sqrt{3} = 50\sqrt{3}$ $= 50 \times 1.73 = 86.5$ $\therefore$ Distance between the ships is $86.5$ m
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