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Question
From a light house 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 light house is h metres, the distance between the ships is
Options
\[\left( \sqrt{3} + 1 \right) \text{ h metres }\]
\[\left( \sqrt{3} - 1 \right) \text{ h metres }\]
\[\sqrt{3} \text{ h metres }\]
\[1 + \left( 1 + \frac{1}{\sqrt{3}} \right) \text{ h metres }\]
Solution
Let the height of the light house AB be h meters
Given that: angle of depression of ship are`∠C=30` and`∠D=45°`
Distance of the ship C =`BC=x` and distance of the ship D =`BD=y`
Here, we have to find distance between the ships.
So we use trigonometric ratios.
In a triangle,`ABC`
`⇒ tan C=(AB)/(BC)`
`⇒ tan 30°=h/x`
`⇒1/sqrt3=h/x`
`⇒ x=sqrt(3h)`
Again in a triangle ABD,
`tan D=(AB)/(BD)`
`⇒ tan D= (AB)/(BD)`
`⇒ tan 45°=h/y`
`⇒ 1=h/y`
`⇒ y=h`
Now, distance between the ships `=x+y=sqrt3h+h(sqrt3+1)h`
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