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

विकल्प

• h tan (45° + θ)

• h cot (45° − θ)

• h tan (45° − θ)

• h cot (45° + θ)

Answer 1

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)] `