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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

$i n t r i a n g l e A \overset{▴}{B C} : ∠ A = 72^{∙}, B C = 10 .$ $D a n d E, c a n m o v e o n B C, b u t s u c h$ $t h a t i n A \overset{▴}{D E}, a l w a y s A D = D E .$ $f i n d : ∠ D A E, w h e n a r e a o f A \overset{▴}{D E},$ $m e e t s m a x i m u m v a l v e .$
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jun/18

$e x c e l l e n t q u e s t i o n . . . l e t t h e c e l l o f b r a i n a c t i v e$
	Answered by ajfour last updated on 16/Jun/18

	$l e t ∠ D A E = θ; ∠ A C B = ϕ$ $i n △ A B C$ $\frac{a}{\sin α} = \frac{b}{\sin (α + ϕ)} . . . . (i)$ $i n △ A C E$ $\frac{b}{\sin θ} = \frac{2 ρ \cos θ}{\sin ϕ} . . . . (i i)$ $S_{A D E} = S = \frac{ρ}{2} (b \sin ϕ) . . . . (i i i)$ $u s i n g (i i) i n (i i i ($ $S = \frac{b^{2} \sin^{2} ϕ}{2 \sin 2 θ}$ $u s i n g (i)$ $S (θ, ϕ) = (\frac{a^{2} \sin^{2} (α + ϕ)}{\sin^{2} α}) (\frac{\sin^{2} ϕ}{2 \sin 2 θ})$ $W h e n S i s m a x i m u m$ $\frac{\partial S}{\partial θ} = 0; \Rightarrow$ $\frac{\partial S}{\partial θ} = \frac{a^{2} \sin^{2} (α + ϕ) \sin^{2} ϕ}{2 \sin^{2} α} (- \frac{2 \cos 2 θ}{\sin^{2} 2 θ}) = 0$ $\Rightarrow \cos 2 θ = 0 o r θ = \frac{π}{4} .$ $\frac{\partial S}{\partial ϕ} = \frac{a^{2}}{2 \sin^{2} α \sin 2 θ} [\sin 2 ϕ \sin^{2} (α + ϕ)$ $+ \sin^{2} ϕ \sin (2 α + 2 ϕ)] = 0$ $\Rightarrow \cos ϕ [1 - \cos (2 α + 2 ϕ)]$ $= - \sin ϕ \sin (2 α + 2 ϕ)$ $o r \cos ϕ = \cos (2 α + 3 ϕ)$ $\Rightarrow 2 π - ϕ = 2 α + 3 ϕ$ $ϕ = \frac{π}{2} - \frac{α}{2}$ $S_{m a x} = \frac{a^{2} \sin^{2} (\frac{π}{2} + \frac{α}{2}) \sin^{2} (\frac{π}{2} - \frac{α}{2})}{2 \sin^{2} α}$ $= \frac{a^{2}}{4} \tan^{2} (\frac{α}{2}) .$ $E l i e s o n B C p r o d u c e d t o w a r d s$ $r i g h t . ∠ A D E = \frac{π}{2} (y e s)$ $∠ D A E = \frac{π}{4} t o m e e t s u c h a$ $c o n d i t i o n .$
	Commented by behi83417@gmail.com last updated on 16/Jun/18

	$i f ∠ D A E = \frac{π}{4} \Rightarrow ∠ A D E = \frac{π}{2} .$ $d o y o u h a v e a n y i d e a s i r ?$
	Commented by behi83417@gmail.com last updated on 15/Jun/18

	$t h a n k y o u s o m u c h d e a r A j f o u r .$ $c a n y o u c a l c u l a t e m a x i m u m a r e a o f A \overset{▴}{D E}$ $w i t h t h i s g i v e n p a r t s o f A \overset{▴}{B C} ?$ $w h a t i s y o u r i d e a w h e n : A D = A E,$ $w i t h s a m e c o n d i t i o n ?$
	Commented by ajfour last updated on 15/Jun/18

	$M y a n s w e r i s p o s t e d S i r .$
	Commented by behi83417@gmail.com last updated on 15/Jun/18

	$s i r A j f o u r! i m w a i t i n g f o r y o u r a n s w e r .$ $w h y d o y o u d e l e t e y o u r n i c e f i g u r e ?$
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