Two ships are there in the sea on either side of a light house in such a way that the ships and the light house are in the same straight line. The angles of depression of two ships as observed from the top of the light house are $60^{o}$ and $45^{o}$. If the height of the light house is $200\ m$, find the distance between the two ships.
Given: A light house of $200\ m$ height and from the top of the light house angle of depression of two ships on either side $60^{o} $ and $45^{o}$.
To do: To find the distance between the ships.
Solution:
Let A and B are two ships and OD is the light house, d is the distance between the two ships.
Let us assume the distance of one of the ships from the lighthouse is x meters
, then the distance of the other ship from the light house is $( d-x)$ meter
In right-angled $\vartriangle ADO$, we have
$tan45^{o}=\frac{OD}{AD}$
$\Rightarrow 1=\frac{200}{x}$
$\Rightarrow x=200\ m$
And in right tringle $\vartriangle ODB$,
$tan60^{o}=\frac{OD}{d-x}$
$\Rightarrow \sqrt{3} =\frac{200}{d-200}$
$\Rightarrow \sqrt{3}( d-200) =200$
$d\sqrt{3} -200\sqrt{3} =200$
$\Rightarrow d\sqrt{3} =200+200\sqrt{3}$
$\Rightarrow d=200\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)$
$\Rightarrow d=200\times 1.58$
$\Rightarrow d=316\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( approximately)$
Thus the distance between the two ships is $316\ m$ approximately.
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