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Question Number 182149 by Acem last updated on 05/Dec/22

Commented by Acem last updated on 05/Dec/22

$F i n d α$
	Answered by HeferH last updated on 05/Dec/22

	$B y c o n s t r u c t i o n :$ $40 ° - α = α - 20 °$ $α = 30 °$
	Commented by Acem last updated on 05/Dec/22

	$30 i s c o r r e c t, b u t h o w ?$
	Answered by HeferH last updated on 05/Dec/22

	Commented by Acem last updated on 05/Dec/22

	$T h a n k s! i^{'} l l c h e c k y o u r m e t h o d v e r y s o o n$ $T h a n k y o u$
	Answered by a.lgnaoui last updated on 06/Dec/22
	$△ABD ∡ABD=180−(80+20)=80 ⇒AD=BD AB^2 =2AD^2 (1−cos 20) AD^2 =((AB^2 )/(2(1−cos 20))) (1) △ACD CD=AB⇒ AD^2 =AC^2 +AB^2 +2AB×ACcos α △ABC ABsin 80=ACsin α (2) AD^2 =((AB^2 sin^2 80)/(sin^2 α))+AB^2 +((2AB^2 )/(sin α))×sin 80cos α =AB^2 (((sin^2 80)/(sin^2 α))+((2sin 80cos α)/(sin α))+1) =AB^2 [(((sin^2 80)/(sin^2 α))+((2sin 80cos α)/(sin α))+1]=((AB^2 )/(2(1−cos 20))) (1/(2(1−cos 20)))=((sin^2 80(1+tan^2 α))/(tan^2 α))+((2sin 80)/(tan α))+1) (1/(2(1−cos 20)))=[((sin^2 80(1+tan^2 α)+2sin 80tan α+tan^2 α)/(tan^2 α))] =[(((sin^2 80+1)tan^2 α+2sin 80tan α+sin^2 80)/(tan^2 α))] posons t=tan α t^2 =2(1−cos 20)[(sin^2 80+1)t^2 +2sin 80t+sin^2 80] 0=[2sin^2 80(1−cos 20)−2cos 20+1]t^2 +(2sin 80)t+sin^2 80 t^2 +((2sin 80)/(1+2sin^2 80(1−cos 20)−2cos 20))t+((sin^2 80)/(1+2sin^2 80(1−cos 20)−2cos 20))=0 sin 80=0,98480775 cos 20=0,939692620785 sin^2 80=0,969846310393 t^2 +((1,969615506)/(0,762407463))t+((0,96984631039)/(0,762407463)) t^2 +2,5834158258 t+1,272081764 t=0,662082972 t=3,842665708 α=tan^(−1) (t)={33,5 ; 75,5} Angle α=33,5°$
	$△ ABD ∡ ABD = 180 - (80 + 20) = 80 \Rightarrow AD = BD$ ${AB}^{2} = {2 AD}^{2} (1 - \cos 20)$ ${AD}^{2} = \frac{{AB}^{2}}{2 (1 - \cos 20)} (1)$ $△ ACD CD = AB \Rightarrow {AD}^{2} = {AC}^{2} + {AB}^{2} + 2 AB \times ACcos α$ $△ ABC ABsin 80 = ACsin α (2)$ ${AD}^{2} = \frac{{AB}^{2} \sin^{2} 80}{\sin^{2} α} + {AB}^{2} + \frac{{2 AB}^{2}}{\sin α} \times \sin 80 \cos α$ $= {AB}^{2} (\frac{\sin^{2} 80}{\sin^{2} α} + \frac{2 \sin 80 \cos α}{\sin α} + 1)$ $= {AB}^{2} [(\frac{\sin^{2} 80}{\sin^{2} α} + \frac{2 \sin 80 \cos α}{\sin α} + 1] = \frac{{AB}^{2}}{2 (1 - \cos 20)}$ $\frac{1}{2 (1 - \cos 20)} = \frac{\sin^{2} 80 (1 + \tan^{2} α)}{\tan^{2} α} + \frac{2 \sin 80}{\tan α} + 1)$ $\frac{1}{2 (1 - \cos 20)} = [\frac{\sin^{2} 80 (1 + \tan^{2} α) + 2 \sin 80 \tan α + \tan^{2} α}{\tan^{2} α}]$ $= [\frac{(\sin^{2} 80 + 1) \tan^{2} α + 2 \sin 80 \tan α + \sin^{2} 80}{\tan^{2} α}]$ $p o s o n s t = \tan α$ $t^{2} = 2 (1 - \cos 20) [(\sin^{2} 80 + 1) t^{2} + 2 \sin 80 t + \sin^{2} 80]$ $0 = [2 \sin^{2} 80 (1 - \cos 20) - 2 \cos 20 + 1] t^{2} + (2 \sin 80) t + \sin^{2} 80$ $t^{2} + \frac{2 \sin 80}{1 + 2 \sin^{2} 80 (1 - \cos 20) - 2 \cos 20} t + \frac{\sin^{2} 80}{1 + 2 \sin^{2} 80 (1 - \cos 20) - 2 \cos 20} = 0$ $\sin 80 = 0, 98480775 \cos 20 = 0, 939692620785$ $\sin^{2} 80 = 0, 969846310393$ $t^{2} + \frac{1, 969615506}{0, 762407463} t + \frac{0, 96984631039}{0, 762407463}$ $t^{2} + 2, 5834158258 t + 1, 272081764$ $t = 0, 662082972$ $t = 3, 842665708$ $α = \tan^{- 1} (t) = {33, 5; 75, 5}$ $A n g l e α = 33, 5 °$
	Commented by a.lgnaoui last updated on 06/Dec/22

	$l o o k a t g r a p h e$ $α = 33, 5$
	Commented by a.lgnaoui last updated on 06/Dec/22

	Commented by Acem last updated on 06/Dec/22

	$M o n s i e u r I g n a o u i, m e r c i b e a u c o u p p o u r {p a r t a g e}^{'}$ $I l e s t p o s s i b l e {q u}^{'} i l y a i t e u u n e e r r e u r l o r s$ $d u c a l c u l, l^{'} a n g l e e s t 30 °$ $V o u s p o u v e z {v e}^{'} r i f i e r l a s o l u t i o n s e l o n l a$ ${m e}^{'} t h o d e E u c l i d i e n n e c i - d e s s o u s$
	Commented by a.lgnaoui last updated on 06/Dec/22
	$apres verification de calculs t^2 +((4sin 80(1−cos 20))/(2sin^2 80(1−cos 20)−2cos 20+1))t+((2sin^2 80(1−cos 20))/(2sin^2 80(1−cos 20)−2cos 20+1)) t^2 −0,3115980757584 t−0,1534321003521 {_(t2=−0,5315043867363) ^(t_1 = +0,5773502691266) soit α=29,99999 donc α=30 est solution autre=−28(exclu)$
	$a p r e s v e r i f i c a t i o n d e c a l c u l s$ $t^{2} + \frac{4 \sin 80 (1 - \cos 20)}{2 \sin^{2} 80 (1 - \cos 20) - 2 \cos 20 + 1} t + \frac{2 \sin^{2} 80 (1 - \cos 20)}{2 \sin^{2} 80 (1 - \cos 20) - 2 \cos 20 + 1}$ $t^{2} - 0, 3115980757584 t - 0, 1534321003521$ ${_{t 2 = - 0, 5315043867363}^{t_{1} = + 0, 5773502691266}$ $s o i t α = 29, 99999$ $d o n c α = 30 e s t s o l u t i o n$ $a u t r e = - 28 (e x c l u)$
	Answered by Acem last updated on 06/Dec/22

	Commented by Acem last updated on 06/Dec/22

	$M o n s i e u r I g n a o u i$ $L e s d e u x t r i a n g l e s A C D, P A B s o n t c o n g r u s$
	Commented by a.lgnaoui last updated on 06/Dec/22

	$j e v e r i f i e l e s c a l c u l s$
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