Question

The elevation of a tower at a station A due north of it is $\alpha$ and at a station B due west of A is $\beta$. Prove that the height of the tower is $\dfrac{AB~sin\alpha ~ sin\beta}{\sqrt{sin^2\alpha-sin^2\beta}}$.


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★ Source : RD Sharma [ClassX CBSE]
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Answer 1

Basic Concept Used :-

$\boxed{ \bf \: {cot}^{2}x = {cosec}^{2}x - 1}$

$\boxed{ \bf \:cosecx = \dfrac{1}{sinx}}$

$\boxed{ \bf \:cotx \: = \: \dfrac{Adjacent \: Side}{Opposite \: Side}}$

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

Let OC represents the tower of height 'h' units.

Let A be any point in North of tower OC

Let B be any point in West of tower OC.

Let OA = 'x' units and OB = 'y' units.

Now,

It is given that,

angle of elevation of tower at station A, due north i.e.

$\rm :\longmapsto\:\angle OAC \: = \: \alpha$

and

angle of elevation of tower at station A, due west i.e

$\rm :\longmapsto\:\angle OBC \: = \: \beta$

Now,

$\rm :\longmapsto\:{ \sf \:In \: \triangle \:OAC}$

$\rm :\longmapsto\:cot \alpha = \dfrac{OA}{OC}$

$\rm :\longmapsto\:cot \alpha = \dfrac{x}{h}$

$\bf\implies \:x = h \: cot \alpha - - - (1)$

Now,

$\rm :\longmapsto\:{ \sf \:In \: \triangle \:OBC}$

$\rm :\longmapsto\:cot \beta = \dfrac{OB}{OC}$

$\rm :\longmapsto\:cot \beta = \dfrac{y}{h}$

$\bf\implies \:y = \: h \: cot \beta - - - (2)$

Now,

$\rm :\longmapsto\:In \: \triangle \: OAB$

Using Pythagoras Theorem,

$\rm :\longmapsto\: {AB}^{2} = {y}^{2} - {x}^{2}$

On substituting the values of x and y from above equations we get,

$\rm :\longmapsto\: {AB}^{2} = {(h \: cot \beta )}^{2} - {(h \: cot \alpha )}^{2}$

$\rm :\longmapsto\: {AB}^{2} = {h}^{2} { (cot \beta)}^{2} - {h}^{2} {(cot \alpha)}^{2}$

$\rm :\longmapsto\: {AB}^{2} = {h}^{2} {cot}^{2} \beta -{h}^{2} {cot}^{2} \alpha$

$\rm :\longmapsto\: {AB}^{2} = {h}^{2}( {cot}^{2} \beta -{cot}^{2} \alpha)$

$\rm :\longmapsto\: {AB}^{2} = {h}^{2} \bigg(( {cosec}^{2} \beta - 1) - \: ({cosec}^{2} \alpha \: - 1) \bigg)$

$\rm :\longmapsto\: {AB}^{2} = {h}^{2}( {cosec}^{2} \beta - {cosec}^{2} \alpha )$

$\rm :\longmapsto\: {AB}^{2} = {h}^{2}\bigg(\dfrac{1}{ {sin}^{2}\beta} - \dfrac{1}{ {sin}^{2} \alpha}\bigg)$

$\rm :\longmapsto\: {AB}^{2} = {h}^{2}\bigg(\dfrac{ {sin}^{2} \alpha - {sin}^{2} \beta }{ {sin}^{2}\beta \: {sin}^{2} \alpha } \bigg)$

$\rm :\longmapsto\: {h}^{2} = \dfrac{ {AB}^{2} {sin}^{2} \alpha \: {sin}^{2} \beta }{ {sin}^{2} \alpha - {sin}^{2} \beta }$

$\bf\implies \:h \: = \: \dfrac{AB \: sin \alpha \: sin \beta }{ \sqrt{{sin}^{2} \alpha - {sin}^{2} \beta} }$

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

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 is in the attachment