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Question
The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 100 m apart. the height of the light house is
Options
\[\frac{50}{\sqrt{3 + 1}} m\]
\[\frac{50}{\sqrt{3 - 1}} m\]
\[50 \left( \sqrt{3} - 1 \right) m\]
\[50 \left( \sqrt{3} + 1 \right) m\]
Solution
Let AB= h be the light house.
The given situation can be represented as,
It is clear that`∠C=45°` and `∠D=30°`
Again, let`BC=x` and`CD=100` m is given.
Here, we have to find the height of light house.
So we use trigonometric ratios.
In a triangle,`ABC`
`⇒ tan C=AB/BC`
`⇒ tan 45°=h/x`
`⇒ 1=h/x`
`⇒h=x`
Again in a triangle ABD,
`⇒ tan D=(AB)/(BC+CD)`
`⇒ tan 30°= h/(x+100)`
`⇒1/sqrt3=h/(x+100)`
`⇒sqrt3h=x+100`
Put `x=h`
`⇒ sqrt3h=h+100`
`⇒ h(sqrt3-1)=100`
`⇒h=100/(sqrt3-1)=100`
`⇒ h=100/(sqrt3-1)`
`⇒ h=100/(sqrt3-1) xx (sqrt3+1)/(sqrt3+1)`
`⇒ h=50(sqrt3+1)`
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