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Question Number 29107 by ajfour last updated on 04/Feb/18

Commented by ajfour last updated on 04/Feb/18

Find blue area in terms of a,𝛉,𝛗 .

Findblueareaintermsofa,θ,ϕ.

Answered by mrW2 last updated on 04/Feb/18

let′s take D as origin and DC as x−axis. y_G =a tan θ (y_F /(tan φ))+(y_F /(tan θ))=2a ⇒y_F =((2a)/((1/(tan φ))+(1/(tan θ))))=((2a)/(((cos φ)/(sin φ))+((cos θ)/(sin θ))))=((2a sin φ sin θ)/(sin (φ+θ))) ⇒x_F =a−(y_F /(tan θ))=a−((2a sin φ cos θ)/(sin (φ+θ)))=−((a sin (φ−θ))/(sin (φ+θ))) similarly, ⇒y_A =((2a sin φ sin 2θ)/(sin (φ+2θ))) ⇒x_A =−((a sin (φ−2θ))/(sin (φ+2θ))) Eqn. of CF: (y/(x−a))=−tan θ ⇒y+tan θ x=a tan θ Eqn. of DA: (y/x)=(((2a sin φ sin 2θ)/(sin (φ+2θ)))/(−((a sin (φ−2θ))/(sin (φ+2θ)))))=−((2 sin φ sin 2θ)/(sin (φ−2θ))) ⇒y=−((2 sin φ sin 2θ)/(sin (φ−2θ)))x ⇒[−((2 sin φ sin 2θ)/(sin (φ−2θ)))+tan θ]x_H =a tan θ ⇒x_H =−((a tan θ)/(((2 sin φ sin 2θ)/(sin (φ−2θ)))−tan θ)) ⇒x_H =−(a/(((4 sin φ cos^2 θ)/(sin (φ−2θ)))−1)) A_(Blue) =(1/2)y_D ∣x_H ∣=(a^2 /2)×((tan θ)/(((4 sin φ cos^2 θ)/(sin (φ−2θ)))−1)) ⇒A_(Blue) =(a^2 /2)×((sin (φ−2θ) tan θ)/(2sin φ +sin (φ+2θ)))

let′stakeDasoriginandDCasx−axis.yG=atanθyFtanϕ+yFtanθ=2a⇒yF=2a1tanϕ+1tanθ=2acosϕsinϕ+cosθsinθ=2asinϕsinθsin(ϕ+θ)⇒xF=a−yFtanθ=a−2asinϕcosθsin(ϕ+θ)=−asin(ϕ−θ)sin(ϕ+θ)similarly,⇒yA=2asinϕsin2θsin(ϕ+2θ)⇒xA=−asin(ϕ−2θ)sin(ϕ+2θ)Eqn.ofCF:yx−a=−tanθ⇒y+tanθx=atanθEqn.ofDA:yx=2asinϕsin2θsin(ϕ+2θ)−asin(ϕ−2θ)sin(ϕ+2θ)=−2sinϕsin2θsin(ϕ−2θ)⇒y=−2sinϕsin2θsin(ϕ−2θ)x⇒[−2sinϕsin2θsin(ϕ−2θ)+tanθ]xH=atanθ⇒xH=−atanθ2sinϕsin2θsin(ϕ−2θ)−tanθ⇒xH=−a4sinϕcos2θsin(ϕ−2θ)−1ABlue=12yD∣xH∣=a22×tanθ4sinϕcos2θsin(ϕ−2θ)−1⇒ABlue=a22×sin(ϕ−2θ)tanθ2sinϕ+sin(ϕ+2θ)

Commented by ajfour last updated on 04/Feb/18

thanks Sir! I could obtain, taking D as origin A=((a^2 tan^2 𝛉)/(2[tan 𝛉+(((2tan 𝛗 tan 2𝛉)/(tan 2𝛉−tan 𝛗)))])) had not rearranged.

thanksSir!Icouldobtain,takingDasoriginA=a2tan2θ2[tanθ+(2tanϕtan2θtan2θ−tanϕ)]hadnotrearranged.

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