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?]](https://www.tinkutara.com/question/Q49731.png)
$$\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{vertex}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{equilateral}}\:\boldsymbol{\mathrm{triangle}}\:\boldsymbol{\mathrm{lies}} \\ $$$$\boldsymbol{\mathrm{on}}\:\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{vertex}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{square}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{two}} \\ $$$$\boldsymbol{\mathrm{anothers}}\:\boldsymbol{\mathrm{lie}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{opposite}}\:\boldsymbol{\mathrm{sides}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{square}} \\ $$$$\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{triangle}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{maximum}} \\ $$$$\boldsymbol{\mathrm{area}}. \\ $$$$\boldsymbol{\mathrm{find}}: \\ $$$$\mathrm{1}.\boldsymbol{\mathrm{ratio}}\:\boldsymbol{\mathrm{of}}:\:\:\:\:\:\frac{\boldsymbol{\mathrm{square}}\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{side}}}{\boldsymbol{\mathrm{triangle}}\:\:\:\:\:\:\:\boldsymbol{\mathrm{side}}} \\ $$$$\mathrm{2}.\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{between}}\:\boldsymbol{\mathrm{square}}\:\boldsymbol{\mathrm{side}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{triangle}} \\ $$$$\boldsymbol{\mathrm{side}}.\left[\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{additional}}\:\boldsymbol{\mathrm{data}}?\right] \\ $$
Answered by mr W last updated on 10/Dec/18
$${there}\:{is}\:{only}\:{one}\:{such}\:{equilateral} \\ $$$${triangle}. \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:{let}\:\theta={angle}\:{between}\:{square}\:{side} \\ $$$${and}\:{triangle}\:{side}, \\ $$$$\mathrm{2}\theta+\mathrm{60}°=\mathrm{90}° \\ $$$$\Rightarrow\theta=\mathrm{15}° \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\: \\ $$$$\frac{{square}\:{side}}{{triangle}\:{side}}=\mathrm{cos}\:\theta=\mathrm{cos}\:\mathrm{15}°=\sqrt{\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{30}°}{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}}{\mathrm{4}}\approx\mathrm{0}.\mathrm{966} \\ $$
Commented by behi83417@gmail.com last updated on 10/Dec/18
$${thanks}\:{in}\:{advance}\:{dear}\:{master}. \\ $$$${what}\:{is}\:{your}\:{idea}\:{if}\:{triangle}\:{be} \\ $$$${isoscale}? \\ $$
Commented by mr W last updated on 10/Dec/18
$${if}\:{the}\:{triangle}\:{should}\:{only}\:\:{be}\:{isosceles},\:{the} \\ $$$${largest}\:{triangle}\:{is}\:{obviously}\:{half}\:{of} \\ $$$${the}\:{square}. \\ $$
Commented by behi83417@gmail.com last updated on 10/Dec/18
$${thanks}\:{dear}\:{master}. \\ $$