The angle of elevation of a cloud from a point h metres above a lake is β. The angle of depression of its reflection in the lake is 45°. The height of location of the cloud from the lake is

#### Options

`("h"(1 + tan beta))/(1 - tan beta)`

`("h"(1 - tan beta))/(1 + tan beta)`

h tan(45° − β)

none of these

#### Solution

**`("h"(1 + tan beta))/(1 - tan beta)`**

**Explanation;**

Hint:

Consider the height of the cloud PC b x

PD = x − h

Let BC be y

In the right ΔADP, tan β = `"PD"/"AD"`

tan β = `(x - "h")/y`

⇒ y = `(x - "h")/(tan beta)` ...(1)

In the right ΔAQD, tan 45° = `"DQ"/"AD"`

1 = `(x + "h")/y`

⇒ y = x + h ...(2)

From (1) and (2) we get,

`(x - "h")/(tan beta)` = x + h

⇒ (x + h) tan β = x − h

⇒ x tan β + h tan β = x − h

h + h tan β = x − x tan β

⇒ h(1 + tan β) = x(1 − tan β)

x = `("h"(1 + tan beta))/((1 - tan beta))`