Question

MATHS - CLASS X

$The \: angle \: of \: elevation \: of \: a \: cloud \\ from \: a \: point \: h \: metres \: above \: the \\ surface \: of \: a \: lake \: is \: \theta \: and \: the \: angle \\ of \: depression \: of \: its \: reflection \\ in \: the \: lake \: is \: \alpha . \: Prove \: that \: the \\ height \: of \: the \: cloud \: above \: the \: lake \\ is \: h( \frac{ \tan\alpha + \tan \theta }{ \tan\alpha - \tan \theta } ).$

Answer 1

Let C be the Position of the Cloud

A-B Line Represents the Line which is 'h' metres above the Lake.

A is the Point from which the Cloud is Observed and θ is the Angle of Elevation of the Cloud viewed from Point A

C'' is the Reflection of the Cloud seen from Point A

α is the Angle of Depression of the Cloud reflection viewed from Point A

LK is the Surface of the Lake

z is the Height of the Cloud from the Point where it is Viewed (i.e.) AB Line.

From the Figure :

$Tan(theta) = \frac{z}{AB}$

⇒ $AB = \frac{z}{Tan(theta)}$

In the Similar way :

$Tan\alpha = \frac{2h + z}{AB}$

⇒ $AB = \frac{2h + z}{Tan\alpha}$

Equating Both 'AB's

⇒ $\frac{z}{Tan(theta)} = \frac{2h + z}{Tan\alpha}$

⇒ z × Tanα = 2h × Tanθ + z × Tanθ

⇒ z(Tanα - Tanθ) = 2hTanθ

⇒ $z = \frac{2hTan(theta)}{Tan\alpha - Tan(theta)}$

But the Height of the Cloud above the Lake is z + h

⇒ $z + h = \frac{2hTan(theta)}{Tan\alpha - Tan(theta)} + h$

⇒ $z + h = \frac{2hTan(theta) - hTan(theta) + hTan\alpha}{Tan\alpha - Tan(theta)}$

⇒ $z + h = \frac{h(Tan\alpha + Tan(theta))}{Tan\alpha - Tan(theta)}$