Question Number 37582 by behi83417@gmail.com last updated on 15/Jun/18
Commented by ajfour last updated on 15/Jun/18
Commented by behi83417@gmail.com last updated on 15/Jun/18
$${in}\:{triangle}\:{A}\overset{\blacktriangle} {{B}C}:\angle{A}=\mathrm{72}^{\bullet} ,{BC}=\mathrm{10}. \\ $$$${D}\:{and}\:{E},{can}\:{move}\:{on}\:{BC},{but}\:{such}\: \\ $$$${that}\:{in}\:{A}\overset{\blacktriangle} {{D}E},{always}\:{AD}={DE}. \\ $$$${find}\::\angle{DAE}\:,{when}\:{area}\:{of}\:{A}\overset{\blacktriangle} {{D}E}, \\ $$$${meets}\:{maximum}\:{valve}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jun/18
$${excellent}\:{question}…{let}\:{the}\:{cell}\:{of}\:{brain}\:{active} \\ $$
Answered by ajfour last updated on 16/Jun/18
![let ∠DAE=θ ; ∠ACB=φ in △ABC (a/(sin α))=(b/(sin (α+φ))) ....(i) in △ ACE (b/(sin θ))=((2ρcos θ)/(sin φ)) ....(ii) S_(ADE) =S =(ρ/2)(bsin φ) ....(iii) using (ii) in (iii( S =((b^2 sin^2 φ)/(2sin 2θ)) using (i) S(θ,φ)=(((a^2 sin^2 (α+φ))/(sin^2 α)))(((sin^2 φ)/(2sin 2θ))) When S is maximum (∂S/∂θ) =0 ; ⇒ (∂S/∂θ)= ((a^2 sin^2 (α+φ)sin^2 φ)/(2sin^2 α))(− ((2cos 2θ)/(sin^2 2θ)))=0 ⇒ cos 2θ=0 or 𝛉=(𝛑/4) . (∂S/∂φ)=(a^2 /(2sin^2 αsin 2θ))[sin 2φsin^2 (α+φ) +sin^2 φsin (2α+2φ)]=0 ⇒ cos φ[1−cos (2α+2φ)] = − sin φsin (2α+2φ) or cos φ= cos (2α+3φ) ⇒ 2π−φ = 2α+3φ 𝛗 = (𝛑/2)−(𝛂/2) S_(max) =((a^2 sin^2 ((π/2)+(α/2))sin^2 ((π/2)−(α/2)))/(2sin^2 α)) =(a^2 /4)tan^2 ((α/2)) . E lies on BC produced towards right. ∠ADE = (π/2) (yes) ∠DAE = (π/4) to meet such a condition.](https://www.tinkutara.com/question/Q37612.png)
$${let}\:\angle{DAE}=\theta\:;\:\:\angle{ACB}=\phi \\ $$$${in}\:\bigtriangleup{ABC} \\ $$$$\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{sin}\:\left(\alpha+\phi\right)}\:\:\:\:\:\:….\left({i}\right) \\ $$$${in}\:\bigtriangleup\:{ACE} \\ $$$$\frac{{b}}{\mathrm{sin}\:\theta}=\frac{\mathrm{2}\rho\mathrm{cos}\:\theta}{\mathrm{sin}\:\phi}\:\:\:\:\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$${S}_{{ADE}} ={S}\:=\frac{\rho}{\mathrm{2}}\left({b}\mathrm{sin}\:\phi\right)\:\:\:\:\:….\left({iii}\right) \\ $$$${using}\:\left({ii}\right)\:{in}\:\left({iii}\left(\right.\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{S}\:=\frac{{b}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \phi}{\mathrm{2sin}\:\mathrm{2}\theta} \\ $$$${using}\:\left({i}\right) \\ $$$$\:\:\:{S}\left(\theta,\phi\right)=\left(\frac{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \left(\alpha+\phi\right)}{\mathrm{sin}\:^{\mathrm{2}} \alpha}\right)\left(\frac{\mathrm{sin}\:^{\mathrm{2}} \phi}{\mathrm{2sin}\:\mathrm{2}\theta}\right) \\ $$$${When}\:{S}\:{is}\:{maximum} \\ $$$$\:\:\:\frac{\partial{S}}{\partial\theta}\:=\mathrm{0}\:;\:\Rightarrow\: \\ $$$$\:\:\:\frac{\partial{S}}{\partial\theta}=\:\:\frac{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \left(\alpha+\phi\right)\mathrm{sin}\:^{\mathrm{2}} \phi}{\mathrm{2sin}\:^{\mathrm{2}} \alpha}\left(−\:\frac{\mathrm{2cos}\:\mathrm{2}\theta}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\theta}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{cos}\:\mathrm{2}\theta=\mathrm{0}\:\:\:{or}\:\:\:\boldsymbol{\theta}=\frac{\boldsymbol{\pi}}{\mathrm{4}}\:. \\ $$$$\frac{\partial{S}}{\partial\phi}=\frac{{a}^{\mathrm{2}} }{\mathrm{2sin}\:^{\mathrm{2}} \alpha\mathrm{sin}\:\mathrm{2}\theta}\left[\mathrm{sin}\:\mathrm{2}\phi\mathrm{sin}\:^{\mathrm{2}} \left(\alpha+\phi\right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{sin}\:^{\mathrm{2}} \phi\mathrm{sin}\:\left(\mathrm{2}\alpha+\mathrm{2}\phi\right)\right]=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{cos}\:\phi\left[\mathrm{1}−\mathrm{cos}\:\left(\mathrm{2}\alpha+\mathrm{2}\phi\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\:\mathrm{sin}\:\phi\mathrm{sin}\:\left(\mathrm{2}\alpha+\mathrm{2}\phi\right) \\ $$$${or}\:\:\:\mathrm{cos}\:\phi=\:\mathrm{cos}\:\left(\mathrm{2}\alpha+\mathrm{3}\phi\right) \\ $$$$\Rightarrow\:\:\mathrm{2}\pi−\phi\:=\:\mathrm{2}\alpha+\mathrm{3}\phi \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}}−\frac{\boldsymbol{\alpha}}{\mathrm{2}}\: \\ $$$${S}_{{max}} =\frac{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}+\frac{\alpha}{\mathrm{2}}\right)\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2sin}\:^{\mathrm{2}} \alpha} \\ $$$$\:\:\:\:\:=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)\:. \\ $$$${E}\:{lies}\:{on}\:{BC}\:{produced}\:{towards} \\ $$$${right}.\:\:\angle{ADE}\:=\:\frac{\pi}{\mathrm{2}}\:\:\:\left({yes}\right) \\ $$$$\angle{DAE}\:=\:\frac{\pi}{\mathrm{4}}\:\:{to}\:{meet}\:{such}\:{a} \\ $$$${condition}. \\ $$
Commented by behi83417@gmail.com last updated on 16/Jun/18
$${if}\:\angle{DAE}=\frac{\pi}{\mathrm{4}}\Rightarrow\angle{ADE}=\frac{\pi}{\mathrm{2}}. \\ $$$${do}\:{you}\:{have}\:{any}\:{idea}\:{sir}? \\ $$
Commented by behi83417@gmail.com last updated on 15/Jun/18
$${thank}\:{you}\:{so}\:{much}\:{dear}\:{Ajfour}. \\ $$$${can}\:\:{you}\:{calculate}\:{maximum}\:{area}\:{of}\:{A}\overset{\blacktriangle} {{D}E} \\ $$$${with}\:{this}\:{given}\:{parts}\:{of}\:{A}\overset{\blacktriangle} {{B}C}? \\ $$$${what}\:{is}\:{your}\:{idea}\:{when}:\:{AD}={AE}, \\ $$$${with}\:{same}\:{condition}? \\ $$
Commented by ajfour last updated on 15/Jun/18
$${My}\:{answer}\:{is}\:{posted}\:{Sir}. \\ $$
Commented by behi83417@gmail.com last updated on 15/Jun/18
$${sir}\:{Ajfour}!{im}\:{waiting}\:{for}\:{your}\:{answer}. \\ $$$${why}\:{do}\:{you}\:{delete}\:{your}\:{nice}\:{figure}? \\ $$