Question

HCG
38.
If the angle of elevation of a Cloud from a point hm above a
lake is d-and the angle of depression of its reflection in
the lake is B. prove that the distance of the cloud from the
point of observation is ah secar COR)
tanp-tand​

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

Figure is in attachment.

Given :-

• A is the point h m above the lake
• The angle of elevation of the cloud from point A is alpha
• The angle of depression of its reflection in the lake from the point A is beta
• A is the point of observation

To prove:-

$\tt{AB=\dfrac{2\:h\:sec\:\alpha}{tan\:\beta-tan\:\alpha}}$

Proof:-

We know that,

$\huge\leadsto{\boxed{\tt{tan\:\theta=\dfrac{Perpendicular}{Base}}}}$

So, In ∆ ABC,

$\leadsto\tt{tan\:\alpha=\dfrac{x}{AB}}$

$\leadsto\tt{AB=\dfrac{x}{tan\:\alpha}}$.....(1)

And, In ∆ ABD,

$\leadsto\tt{tan\:\beta=\dfrac{x+2h}{AB}}$

$\leadsto\tt{AB=\dfrac{X+2h}{tan\:\beta}}$.....(2)

Now, by (1) and (2),

$\leadsto\tt{AB=\dfrac{x}{tan\:\alpha}=\dfrac{x+2h}{tan\:\beta}}$

$\leadsto\tt{x\:tan\:\alpha=(X+2h)tan\:\alpha}$

$\leadsto\tt{x\:tan\:\alpha=X\:tan\:\alpha+2h\:tan\:\alpha}$

$\leadsto\tt{x\:tan\:\alpha-tan\:\beta=2h\:tan\:\alpha}$

$\leadsto\tt{x(tan\:\beta-tan\:\alpha)=2h\:tan\:\alpha}$

So,

$\leadsto\tt{x=\dfrac{2h\times tan\:\alpha}{tan\:\beta-tan\:\alpha}}$

Now, we know that,

$\huge\leadsto{\boxed{\tt{sin\:\theta=\dfrac{Perpendicular}{Hypertenuse}}}}$

So, In ∆ ABC,

$\leadsto\tt{sin\:\alpha=\dfrac{x}{AC}}$

Finding x,

$\leadsto\tt{x=sin\:\alpha\times AC}$

Now putting the value of x,

$\leadsto\tt{sin\:\alpha\times AC=\dfrac{2h\times tan\:\alpha}{tan\:\beta}}$

Now,

$\leadsto\tt{AC=\dfrac{2h\times \dfrac{\cancel{sin\:\alpha}}{cos\:\alpha}}{tan\:\beta\times \cancel{sin\:\alpha}}}$

Now,

$\huge\leadsto{\boxed{\red{\tt{AB=\dfrac{2\:h\:sec\:\alpha}{tan\:\beta-tan\:\alpha}}}}}$

Hence proved.

$\rule{200}2$

Formulas Used:-

• $\tt{sin\:\theta=\dfrac{Perpendicular}{Hypotenuse}}$
• $\tt{tan\:\theta=\dfrac{Perpendicular}{Base}}$
• $\tt{tan\:\theta=\dfrac{sin\:\theta}{cos\:\theta}}$
• $\tt{sec\:\theta=\dfrac{1}{cos\:\theta}}$

$\rule{200}4$

Answer 2

||✪✪ QUESTION ✪✪||

If the angle of elevation of a cloud from point h metres above a lake is α and the angle of depression of its reflection in the lake is β, prove that the distance of the cloud from the point of observation is :- 2hsecα/(Tanβ - Tanα)

|| ✰✰ ANSWER ✰✰ ||

❁❁ Refer To Image First .. ❁❁

|| ★★ FORMULA USED ★★ ||

• Tan@ = Perpendicular / Base
• Sin@ = Perpendicular / Hypotenuse
• Tan@ = (sin@/cos@)
• (1/cos@) = sec@ .