Question

Questions

A Ladder set against a wall makes an angle  θ with the horizontal. when it's foot is pulled away from the wall through a distance 'p', it slides a distance 'q' down the wall making an angle ϕ with the horizontal. Show that
$\sf \dfrac{p}{q} = \dfrac{cos \theta - cos \phi}{sin \phi - sin \theta}$

Class = 10
Chapter = Some Applications of Trigonometry​

Answer 1

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

Let AC be the ladder such that it top A touches the wall AB and bottom C on the ground.

The ladder is pulled away from the wall through a distance 'p' , so that it top C slides and take position E on wall and D on ground.

So,

• AC = DE = Length of Ladder.

Let

• Length of Ladder be 'z'

• Distance, BC = 'x'

• Distance, BE = 'y'

Given that

• Distance, DC = p

• Distance, AE = q

Now,

$\rm :\longmapsto\:In \: \triangle \: ABC$

$\rm :\longmapsto\: sin\theta \: = \dfrac{AB}{AC}$

$\rm :\longmapsto\: sin\theta \: =\dfrac{q + y}{z} - - - (1)$

and

$\rm :\longmapsto\: cos\theta \: =\dfrac{BC}{AB}$

$\rm :\longmapsto\: cos\theta \: =\dfrac{x}{z} - - -(2)$

Now,

$\rm :\longmapsto\:In \: \triangle \:DBE$

$\rm :\longmapsto\:sin \phi \: = \dfrac{BE}{DE}$

$\rm :\longmapsto\:sin \phi \: = \dfrac{y}{z} - - - (3)$

and

$\rm :\longmapsto\:cos \phi \: = \dfrac{DB}{DE}$

$\rm :\longmapsto\:cos \phi \: = \dfrac{p + x}{z} - - - (4)$

Consider,

$\rm :\longmapsto\:\dfrac{cos\theta \: - \: cos\phi}{sin\phi \: - \: sin\theta}$

On substituting the values of sin ϕ, cos ϕ, sin θ, cos θ,

$\rm \: \: \: = \: \: \dfrac{\dfrac{x}{z} - \dfrac{p + x}{z} }{ \: \: \: \: \dfrac{y}{z} - \dfrac{q + y}{z} \: \: \: \: \: \: }$

$\rm\: \: \: = \: \: \dfrac{\dfrac{ \cancel{x} - \cancel{x} - p}{\cancel{z}} }{ \: \: \: \: \: \: \: \dfrac{\cancel{y} - \cancel{y} - q}{\cancel{z}} \: \: \: \: \: }$

$\rm \: \: \: = \: \: \dfrac{ - \: p}{ - \: q}$

$\rm\: \: \: = \: \: \dfrac{p}{q}$

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

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