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From a lighthouse the angles of depression of two ships on opposite sides of the light house are observed to be 30° and 45°. If the height of the lighthouse is h meters, the distance between the ships is.

A. $(\sqrt{3}+1) h$ m B. $(\sqrt{3}-1) h$ m C. $\sqrt{3} h$ m D. $1+\left(1+\frac{1 }{\sqrt{3}}\right) h$ m Answer: Option A
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Solution(By Apex Team)

From a lighthouse the angles of depression of two ships image Let AB be lighthouse and P and Q are two ships on its opposite sides which form angle of elevation of A as 45° and 30° respectively AB = h Let PB = x and QB = y $\begin{array}{l}\ text{Now in right }\triangle\text{APB}\\ \tan\theta=\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{AB}{PB}\\ \Rightarrow\ tan45^{\circ}=\frac{h}{x}\Rightarrow1=\frac{h}{x}\\ \Rightarrow x=h\ldots\ldots\ldots\ldots(i)\\ \text{Similarly in right}\triangle\text{AQB,}\\ \begin{array}{l}\tan30^{\circ}=\frac{AP}{QB}=\frac{h}{y}\\ \begin {array}{l}\Rightarrow\frac{1}{\sqrt{3}}=\frac{h}{y}\\ \Rightarrow y=\sqrt{3}h\ldots\ldots\ldots\ldots. (ii)\\ \text{ Adding (i) and (ii) }\\ \begin{aligned}\therefore PQ&=x+y\\ &=h+\sqrt{3}h\\ &=(\sqrt{ 3}+1)h\end{aligned}\end{array}\end{array}\end{array}$