Question Number 38032 by behi83417@gmail.com last updated on 20/Jun/18
Commented by behi83417@gmail.com last updated on 21/Jun/18
$$\boldsymbol{{in}}\:\boldsymbol{{triangle}}\:\boldsymbol{{A}}\overset{\blacktriangle} {\boldsymbol{{B}C}}:\angle\boldsymbol{{A}}=\mathrm{72}^{\bullet} ,\boldsymbol{{BC}}=\mathrm{10}. \\ $$$$\boldsymbol{{D}}\:\boldsymbol{{and}}\:\boldsymbol{{E}},\boldsymbol{{can}}\:\boldsymbol{{move}}\:\boldsymbol{{on}}\:\boldsymbol{{BC}},\boldsymbol{{but}}\:\boldsymbol{{such}} \\ $$$$\boldsymbol{{that}},\boldsymbol{{alwyes}}:\boldsymbol{{AD}}=\boldsymbol{{AE}}. \\ $$$$\left.\mathrm{1}\right)\boldsymbol{{evaluate}}\:\angle\boldsymbol{{DAE}},\boldsymbol{{when}}\:\boldsymbol{{area}}\:\boldsymbol{{of}}\: \\ $$$$\boldsymbol{{triangle}}\:\boldsymbol{{D}}\overset{\blacktriangle} {\boldsymbol{{A}E}},\boldsymbol{{meets}}\:\boldsymbol{{maximum}} \\ $$$$\boldsymbol{{valve}}. \\ $$$$\left.\mathrm{2}\right)\boldsymbol{{in}}\:\boldsymbol{{one}}\:\boldsymbol{{position}}\:\boldsymbol{{of}}:\:\boldsymbol{{D\&E}}, \\ $$$$\boldsymbol{{we}}\:\boldsymbol{{have}}:\boldsymbol{{CD}}=\boldsymbol{{EB}},\boldsymbol{{AD}}=\boldsymbol{{AE}}. \\ $$$$\boldsymbol{{find}}:\angle\boldsymbol{{DAE}}\:\boldsymbol{{and}}\:\boldsymbol{{area}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{A}}\overset{\blacktriangle} {\boldsymbol{{D}E}},\boldsymbol{{for}}\:\boldsymbol{{this}}\:\boldsymbol{{spicial}}\:\boldsymbol{{case}}. \\ $$$$\left.\mathrm{3}\right)\:\boldsymbol{{by}}\:\boldsymbol{{changing}}\:\boldsymbol{{position}}\:\boldsymbol{{of}}\:''\boldsymbol{{A}}'', \\ $$$$\boldsymbol{{we}}\:\boldsymbol{{can}}\:\boldsymbol{{make}}\:\boldsymbol{{the}}\:\boldsymbol{{equation}}: \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{CD}}=\boldsymbol{{DA}}=\boldsymbol{{AE}}=\boldsymbol{{EB}}. \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{evalvate}}:\angle\boldsymbol{{DAE}}\:\&\left(\frac{\boldsymbol{{area}}_{\boldsymbol{{D}}\overset{\blacktriangle} {\boldsymbol{{A}E}}} }{\boldsymbol{{area}}_{\boldsymbol{{A}}\overset{\blacktriangle} {\boldsymbol{{B}C}}} }\right). \\ $$
Answered by MrW3 last updated on 25/Jun/18
![let ∠A=α=72°, ∠B=β, ∠C=γ (1) let′s assume β≥γ, ((AB)/(sin γ))=((BC)/(sin α)) ⇒AB=((sin γ)/(sin α))×a=((sin (α+β) a)/(sin α)) max.A_(ΔADE) =AB×cos β×AB×sin β =((a^2 [sin (α+β)]^2 sin 2β)/(2[sin α]^2 )) =(a^2 /2)[cos^2 β+((sin^2 β)/(tan^2 α))]sin 2β (2) β=γ=((π−α)/2) max.A_(ΔADE) =(a^2 /2)[sin^2 (α/2)+((cos^2 (α/2))/(tan^2 α))]sin α (3) I don′t understand what is meant.](https://www.tinkutara.com/question/Q38447.png)
$${let}\:\angle{A}=\alpha=\mathrm{72}°,\:\angle{B}=\beta,\:\angle{C}=\gamma \\ $$$$\left(\mathrm{1}\right) \\ $$$${let}'{s}\:{assume}\:\beta\geqslant\gamma, \\ $$$$\frac{{AB}}{\mathrm{sin}\:\gamma}=\frac{{BC}}{\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow{AB}=\frac{\mathrm{sin}\:\gamma}{\mathrm{sin}\:\alpha}×{a}=\frac{\mathrm{sin}\:\left(\alpha+\beta\right)\:{a}}{\mathrm{sin}\:\alpha} \\ $$$${max}.{A}_{\Delta{ADE}} ={AB}×\mathrm{cos}\:\beta×{AB}×\mathrm{sin}\:\beta \\ $$$$=\frac{{a}^{\mathrm{2}} \left[\mathrm{sin}\:\left(\alpha+\beta\right)\right]^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\beta}{\mathrm{2}\left[\mathrm{sin}\:\alpha\right]^{\mathrm{2}} } \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left[\mathrm{cos}^{\mathrm{2}} \:\beta+\frac{\mathrm{sin}^{\mathrm{2}} \:\beta}{\mathrm{tan}^{\mathrm{2}} \:\alpha}\right]\mathrm{sin}\:\mathrm{2}\beta \\ $$$$\left(\mathrm{2}\right) \\ $$$$\beta=\gamma=\frac{\pi−\alpha}{\mathrm{2}} \\ $$$${max}.{A}_{\Delta{ADE}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left[\mathrm{sin}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}+\frac{\mathrm{cos}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}}{\mathrm{tan}^{\mathrm{2}} \:\alpha}\right]\mathrm{sin}\:\alpha \\ $$$$\left(\mathrm{3}\right) \\ $$$${I}\:{don}'{t}\:{understand}\:{what}\:{is}\:{meant}. \\ $$