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प्रश्न
Two ships are approaching a light-house from opposite directions. The angles of depression of the two ships from the top of the light-house are 30° and 45°. If the distance between the two ships is 100 m, find the height of the light-house. \[[Use \sqrt{3} = 1 . 732]\]
Two ships are approaching a light-house from opposite directions. The angles of depression of the two ships from the top of the light-house are 30° and 45°. If the distance between the two ships is 100 m, find the height of the light house correct to one decimal place.
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उत्तर
Let height of lighthouse (CD) be h meters.
Let A and B be the boats approaching lighthouse.
From figure,
∠CAD = ∠ACX = 30° and ∠CBD = ∠BCY = 45°
In triangle ACD,
\[\Rightarrow \tan 30^\circ = \frac{h}{x}\]
\[\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x}\]
\[\Rightarrow x = h\sqrt{3}.\]
In triangle BCD,
\[\Rightarrow \tan 45^\circ = \frac{h}{y}\]
\[\Rightarrow 1 = \frac{h}{y}\]
⇒ h = y.
Given,
Distance between the two boats is 100 m.
x + y = 100
\[\Rightarrow h\sqrt{3} + h = 100\]
\[\Rightarrow h(\sqrt{3} + 1) = 100\]
\[\Rightarrow h = \frac{100}{\sqrt{3} + 1}\]
\[\Rightarrow h = \frac{100(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)}\]
\[\Rightarrow h = \frac{100(\sqrt{3} - 1)}{3 - 1}\]
\[\Rightarrow h = \frac{100(\sqrt{3} - 1)}{2}\]
\[\Rightarrow h = 50(\sqrt{3} - 1)\]
\[\Rightarrow h = 50(1.732 - 1)\]
\[\Rightarrow h = 50(0.732)\]
\[\Rightarrow h = 36.6\ \text{m}.\]
