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Question
The angle of elevation of a cloud from a point h metre above a lake is θ. The angle of depression of its reflection in the lake is 45°. The height of the cloud is
Options
h tan (45° + θ)
h cot (45° − θ)
h tan (45° − θ)
h cot (45° + θ)
Solution
Let AB be the surface of the lake and p be the point of observation. So AP=h . The given situation can be represented as,
Here,C is the position of the cloud and C' is the reflection in the lake. Then .`CB=C'B`
Let PM be the perpendicular from P on CB. Then `∠CPM=θ` and .`∠C'PM=45°`
Let`CM=x` , `PM=y` , then`CB=x+h` and `C'B=x+h`
Here, we have to find the height of cloud.
So we use trigonometric ratios.
In , `ΔCMP`
`⇒ tan θ=(CM)/(PM)`
`⇒ tanθ=x/y`
`⇒ y= x/tanθ ` (1)
Again in ΔPMC',
`⇒ tan 45°=CM/PM`
`⇒1=(C'B+BM)/(PM)`
`⇒1=(x+h+h)/y`
`⇒y=x+2h`
`⇒ x/tanθ=x+2h` [using (1)]
`⇒ (x-x tanθ)/tan θ=2h`
`⇒ x=(2h tan θ)/((1-tan θ))`
Now,
`⇒ CB=h+x`
`⇒ CB=h+(2h tan θ)/(1-tan θ)`
`⇒ CB=(h(1-tanθ)+2h tanθ)/(1-tan θ)`
`⇒ CB= (h(1+tan θ))/(1-tan θ)=h tan(45°+ θ )` `[tan (A+B)=(tan A+tan B)/(1-tan A tan B)] `
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