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?]


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

$one vertex of a equilateral triangle lies$ $on one vertex of a square and two$ $anothers lie on opposite sides of square$ $such that triangle have the maximum$ $area .$ $find :$ $1 . ratio of : \frac{square side}{triangle side}$ $2 . angle between square side and triangle$ $side . [need additional data ?]$
	Answered by mr W last updated on 10/Dec/18

	$t h e r e i s o n l y o n e s u c h e q u i l a t e r a l$ $t r i a n g l e .$ $2) l e t θ = a n g l e b e t w e e n s q u a r e s i d e$ $a n d t r i a n g l e s i d e,$ $2 θ + 60 ° = 90 °$ $\Rightarrow θ = 15 °$ $1)$ $\frac{s q u a r e s i d e}{t r i a n g l e s i d e} = \cos θ = \cos 15 ° = \sqrt{\frac{1 + \cos 30 °}{2}} = \frac{\sqrt{2} + \sqrt{6}}{4} \approx 0.966$
	Commented by behi83417@gmail.com last updated on 10/Dec/18

	$t h a n k s i n a d v a n c e d e a r m a s t e r .$ $w h a t i s y o u r i d e a i f t r i a n g l e b e$ $i s o s c a l e ?$
	Commented by mr W last updated on 10/Dec/18

	$i f t h e t r i a n g l e s h o u l d o n l y b e i s o s c e l e s, t h e$ $l a r g e s t t r i a n g l e i s o b v i o u s l y h a l f o f$ $t h e s q u a r e .$
	Commented by behi83417@gmail.com last updated on 10/Dec/18

	$t h a n k s d e a r m a s t e r .$
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