Question

Prove that :

$\bigstar \: \sf \dfrac{a}{b} = \dfrac{cos \alpha - cos \beta}{sin \beta - sin \alpha}$
​

Answer 1

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

Let assume that the AB be the ladder whose top B is on the wall and A is on the ground.

As the ladder pulled away from the wall along the ground at a distance 'a' units, it top B slides down 'b' units.

Let ladder take the new position DE, where E is on the wall and D on the ground.

So, AD = a units and BE = b units.

Let assume that AB = DE = l units and CE = y units and AC = x units.

Now, In right angle triangle ABC

$\rm \: sin\alpha = \dfrac{BC}{AB}$

$\rm \: sin\alpha = \dfrac{y + b}{l} - - - (1)$

Also,

$\rm \: cos\alpha = \dfrac{AC}{AB}$

$\rm \: cos\alpha = \dfrac{x}{l} - - - (2)$

Now, In right triangle CDE

$\rm \: sin\beta = \dfrac{CE}{DE}$

$\rm \: sin\beta = \dfrac{y}{l} - - - (3)$

Also,

$\rm \: cos\beta = \dfrac{DC}{DE}$

$\rm \: cos\beta = \dfrac{x + a}{l} - - - (4)$

Now, Subtracting equation (3) from equation (1), we get

$\rm \: sin\alpha - sin\beta$

$\rm \: = \: \dfrac{y + b}{l} - \dfrac{y}{l}$

$\rm \: = \: \dfrac{y + b - y}{l}$

$\rm \: = \: \dfrac{b}{l}$

$\rm\implies \:sin\alpha - sin\beta = \dfrac{b}{l} - - - (5)$

Now, On Subtracting equation (2) from equation (4), we get

$\rm \: cos\beta - cos\alpha$

$\rm \: = \: \dfrac{a + x}{l} - \dfrac{x}{l}$

$\rm \: = \: \dfrac{a + x - x}{l}$

$\rm \: = \: \dfrac{a}{l}$

So,

$\rm\implies \:cos\beta - cos\alpha = \: \dfrac{a}{l} - - - (6)$

Now, on dividing equation (6) by (5), we get

$\rm \: \dfrac{cos\beta - cos\alpha }{sin\alpha - sin\beta } = \dfrac{a}{b}$

can be further rewritten as

$\rm \: \dfrac{ - (cos\alpha - cos\beta ) }{ - (sin\beta - sin\alpha) } = \dfrac{a}{b}$

$\rm\implies \:\boxed{\sf{ \:\rm \: \frac{cos\alpha - cos\beta }{sin\beta - sin\alpha } \: = \: \frac{a}{b} \: \: }}$

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

Given :

According to given figure AB and CD show or describe same ladder

Hence thier length must be equal or same

Assume that :

Consider the length of ladder be h

$\rm \therefore \: AB=CD = h$

So,

$\rm \: In \: \triangle AEB \: ,$

$\rm \implies \: \dfrac{AE}{AB} = \sin( \alpha )$

$\rm \: and$

$\rm \implies \: \dfrac{BE}{AB} = \cos( \alpha )$

Now,

$\rm \implies \: AE= \: h \: \sin( \alpha )$

$\rm \implies \: BE= \: h \: \cos( \alpha )$

So,

$\rm \: In \: \triangle DEC \: ,$

$\rm \implies \: \dfrac{DE}{CD} = \sin( \beta )$

$\rm \implies \: \dfrac{CE}{CD} = \cos( \beta )$

Now,

$\rm \implies \: DE= \: h \: \sin( \beta )$

$\rm \implies \: CE= \: h \: \cos( \beta )$

Here,

$\rm \implies \: \dfrac{a}{b} = \dfrac{BC}{AD}$

$\rm \implies \: \dfrac{a}{b} = \dfrac{CE-BE}{AE-DE}$

$\rm \implies \: \dfrac{a}{b} = \dfrac{(h \: \cos( \beta ) - h \cos( \alpha ) ) }{(h \: \sin( \alpha ) - h \: \sin( \beta ) ) }$

$\rm \implies \: \dfrac{a}{b} = \dfrac{h \: (\cos \beta - \cos\alpha )}{h \: (\sin\alpha - \sin \beta )}$

$\rm \implies \: \dfrac{a}{b} = \dfrac{ \: \cos \beta - \cos\alpha }{\sin\alpha - \sin \beta }$

$\rm \implies \boxed{ \dfrac{a}{b} = \dfrac{ \: \cos \alpha - \cos \beta }{\sin \beta - \sin \alpha }}$

Hence proved