Question

$\mathfrak{\text{MATHS - CLASS X}}$

$\mathbb {\text {APPLICATIONS OF TRIGONOMETRY}}$

As observed from the top of a 75 m lighthouse from the sea - level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

$\mathcal{\text{NO SPAMMING}}$

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Answer 1

$\bf{Answer\colon}$

$We\:have,\\\\Height\:of\:lighthouse\:=\:75\:m\\\\Let\:the\:distance\:between\:the\:two\\ships\:{S}_{1}\:and\:{S}_{2}\:=\:{S}_{1}{S}_{2}\:=\:x\:m\\and\:{S}_{2}H=y\:m\\\\In\:right\:trianlgle\:LH{S}_{2},\:\angle{H}=90^\circ\\\\Therefore\:Tan\:45^\circ=\frac{LH}{{S}_{2}H}\\\\1=\frac{75}{y}\\\\y=75\\\\Again\:in\:right\:triangle\:LH{S}_{1}\:\angle{H}=90^\circ\\\\Therefore\:Tan\:30^\circ=\frac{75\:m}{x+y}\\\\x+y=75\sqrt{3}\\\\x=75\sqrt{3}-75\\\\x=75(\sqrt{3}-1)\:m.\\\\\\Hence,\:the\:distance\:between\\the\:two\:ships\:{S}_{1}{S}_{2}\\=x\\=75(\sqrt{3}-1)\:m.$

Answer 2

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