Question

a question for stars/mods

refer the attachment
answer asap​

Answer 1

Tᴏ Fɪɴᴅ :-

• Distance of the cloud from observation point. (CP)

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ :

Let AB be the surface of lake

Let P be any point 'h' meter above the lake level.

Let cloud is at the point C

Let D be the reflection of cloud in lake so that BD = BC

Now,

• Angle of elevation of the cloud from P be α, and angle of depression from point P of its reflection be β.

Let

• CE = x meter

• PE = y meter

• DE = BD + DE = h + x + h = 2h + x meter

Now,

$\rm :\longmapsto\:In \: \triangle \: CPE$

$\rm :\longmapsto\:tan \alpha = \dfrac{CE}{PE}$

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

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

Now,

$\rm :\longmapsto\:In \: \triangle \: DEP$

$\rm :\longmapsto\:tan \beta = \dfrac{DE}{EP}$

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

$\rm :\longmapsto\:tan \beta = \dfrac{2h + ytan \alpha }{y} \: \: \: \: \{ \: using \: (1) \}$

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

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

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

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

Again,

$\rm :\longmapsto\:In \: \triangle \: CPE$

$\rm :\longmapsto\:sec \alpha = \dfrac{CP}{PE}$

$\rm :\longmapsto\:sec \alpha = \dfrac{CP}{y}$

$\rm :\longmapsto\:CP = ysec \alpha$

$\rm :\longmapsto\:CP = \dfrac{2h \: sec \alpha }{tan \beta - tan \alpha } \: \: \: \: \: \: \: \: \: \{using \: (2) \}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

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

Answer:

The answer of the question is above

Hope it helps you