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Question Number 49731 by behi83417@gmail.com last updated on 09/Dec/18

one vertex of a equilateral triangle lies  on  one vertex of a square and two  anothers lie on opposite sides of square  such that triangle have the maximum  area.  find:  1.ratio of:     ((square          side)/(triangle       side))  2.angle between square side and triangle  side.[need additional data?]

$$\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{vertex}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{equilateral}}\:\boldsymbol{\mathrm{triangle}}\:\boldsymbol{\mathrm{lies}} \\ $$$$\boldsymbol{\mathrm{on}}\:\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{vertex}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{square}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{two}} \\ $$$$\boldsymbol{\mathrm{anothers}}\:\boldsymbol{\mathrm{lie}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{opposite}}\:\boldsymbol{\mathrm{sides}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{square}} \\ $$$$\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{triangle}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{maximum}} \\ $$$$\boldsymbol{\mathrm{area}}. \\ $$$$\boldsymbol{\mathrm{find}}: \\ $$$$\mathrm{1}.\boldsymbol{\mathrm{ratio}}\:\boldsymbol{\mathrm{of}}:\:\:\:\:\:\frac{\boldsymbol{\mathrm{square}}\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{side}}}{\boldsymbol{\mathrm{triangle}}\:\:\:\:\:\:\:\boldsymbol{\mathrm{side}}} \\ $$$$\mathrm{2}.\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{between}}\:\boldsymbol{\mathrm{square}}\:\boldsymbol{\mathrm{side}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{triangle}} \\ $$$$\boldsymbol{\mathrm{side}}.\left[\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{additional}}\:\boldsymbol{\mathrm{data}}?\right] \\ $$

Answered by mr W last updated on 10/Dec/18

there is only one such equilateral  triangle.    2) let θ=angle between square side  and triangle side,  2θ+60°=90°  ⇒θ=15°    1)   ((square side)/(triangle side))=cos θ=cos 15°=(√((1+cos 30°)/2))=(((√2)+(√6))/4)≈0.966

$${there}\:{is}\:{only}\:{one}\:{such}\:{equilateral} \\ $$$${triangle}. \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:{let}\:\theta={angle}\:{between}\:{square}\:{side} \\ $$$${and}\:{triangle}\:{side}, \\ $$$$\mathrm{2}\theta+\mathrm{60}°=\mathrm{90}° \\ $$$$\Rightarrow\theta=\mathrm{15}° \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\: \\ $$$$\frac{{square}\:{side}}{{triangle}\:{side}}=\mathrm{cos}\:\theta=\mathrm{cos}\:\mathrm{15}°=\sqrt{\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{30}°}{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}}{\mathrm{4}}\approx\mathrm{0}.\mathrm{966} \\ $$

Commented by behi83417@gmail.com last updated on 10/Dec/18

thanks in advance dear master.  what is your idea if triangle be  isoscale?

$${thanks}\:{in}\:{advance}\:{dear}\:{master}. \\ $$$${what}\:{is}\:{your}\:{idea}\:{if}\:{triangle}\:{be} \\ $$$${isoscale}? \\ $$

Commented by mr W last updated on 10/Dec/18

if the triangle should only  be isosceles, the  largest triangle is obviously half of  the square.

$${if}\:{the}\:{triangle}\:{should}\:{only}\:\:{be}\:{isosceles},\:{the} \\ $$$${largest}\:{triangle}\:{is}\:{obviously}\:{half}\:{of} \\ $$$${the}\:{square}. \\ $$

Commented by behi83417@gmail.com last updated on 10/Dec/18

thanks dear master.

$${thanks}\:{dear}\:{master}. \\ $$

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