Question

Prove that the distance of the cloud from the point of observation is :


$\boxed {\sf \dfrac{2h sec \alpha }{tan \beta - tan \alpha} }$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

AB be the surface of lake and C be any point of observation above the lake level such that AC = h units.

Let further assume that D be the position of the cloud and F be the reflection of cloud in the lake.

We know, surface level of lake act as a plane mirror. So, distance of the image is equals to distance of the object.

So, DB = BF.

Now, from C, draw CG perpendicular to DF intersecting DF at G.

Let assume that DG = x units

So, DB = DG + GB = x + h units

So, DB = BF = x + h

Let CG = y and CD = z

Now, Consider right-angled triangle CDG

$\rm \: tan\alpha = \dfrac{DG}{CG}$

$\rm \: tan\alpha = \dfrac{x}{y}$

$\rm\implies \:x = y \: tan\alpha - - - (1)$

Now, In right angle triangle CGF

$\rm \: tan\beta = \dfrac{GF}{CG}$

$\rm \: tan\beta = \dfrac{x + 2h}{y}$

$\rm \: ytan\beta = ytan\alpha + 2h$

$\rm \: ytan\beta - ytan\alpha = 2h$

$\rm \: y(tan\beta - tan\alpha) = 2h$

$\rm\implies \:y = \dfrac{2h}{tan\beta - tan\alpha } - - - (2)$

Now, In right-angle triangle CDG

$\rm \: sec\alpha = \dfrac{CD}{CG}$

$\rm \: sec\alpha = \dfrac{z}{y}$

$\rm \: z = y \: sec\alpha$

On substituting the value of y from equation (2), we get

$\rm\implies \:z = \dfrac{2h \: sec\alpha }{tan\beta - tan\alpha }$

Hence, Proved

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

Answer 2

Question:-

If the angle of elevation of a cloud from a point 'h' meters above a lake is 'α' and the angle of depression of its reflection in the lake be 'β'. Prove that the distance of the cloud from the point of observation is

${\sf \large\dfrac{2h \: sec \alpha }{tan \beta - tan \alpha}. }$

Given:-

• The angle of elevation of a cloud from a point 'h' meters above a lake is 'α' and the angle of depression of its reflection in the lake be 'β'.

To Find:-

${\sf \large \dfrac{2h sec \alpha }{tan \beta - tan \alpha}. }$

Solution:-

Let, AB be a lake. The angle of elevation of cloud P at a point A on a height 'h' from the lake is a and the angle of depression of the reflection F is a cloud is.

So,

∠PCE = α and ∠FCE = β

Let, BP = FB = d

So, PE = BP – EB

=> PE = BP – AC = d – h and FE = FB + AC = d + h

Let, CE = x

In △CEF

$\sf \large tan \beta = \frac{EF}{CE}$

$\sf \large \Rightarrow tan \beta = \frac{d + h}{x} \: \: ...(1)$

In △PEC

$\sf \large tan \alpha = \frac{PE}{CE}$

$\sf \large \Rightarrow tan \alpha = \frac{d + h}{x} \: \: ...(2)$

Subtracting Equation (1) and (2) we get,

$\sf \large tan \beta - tan \alpha = \frac{d + h}{x} - \frac{d - h}{x} = \frac{2h}{x} \Rightarrow x$

$\sf \large = \frac{2h}{tan \beta \: tan \alpha} \: \: ...(3)$

In △PEC

$\sf \large cos \alpha = \frac{CE}{PC} = \frac{x}{PC}$

$\sf \large \Rightarrow PC = \frac{c}{cos \alpha} = x \: sec \alpha$

$\sf \large = PC = \frac{2h}{tan \beta \: tan \alpha}$

Answer:-

Hence, the distance of the cloud from the point of observation

${\sf \large= \dfrac{2h \: sec \alpha }{tan \beta - tan \alpha}. }$

Hope you have satisfied. ⚘