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Question Number 188100 by Shrinava last updated on 25/Feb/23

	Answered by a.lgnaoui last updated on 25/Feb/23
	$△ADC ∡CDA=x+18; AC=b CD= ((sin 18)/b)=((sin (x+18))/a) ⇒(a/b)=((sin (x+18))/(sin 18)) (1) △ABC BD=AC=a BC=a+b ((sin(x+18) )/(a+b))=((sin 18)/a) a[sin (x+18)−sin 18]=bsin 18 (a/b)=((sin 18)/(sin (x+18)−sin 18)) (2) (1)=(2)⇒ ((sin (x+18))/(sin 18)) =((sin 18)/(sin (x+18)−sin 18)) posons z=sin (x+18) sin 18=c (z/c) =(c/(z−c))⇒ [z^2 −cz−c^2 =0] △=c^2 +4c { ((z=((c±(√(c^2 +4c^2 )))/2)=((c(1+(√5) ))/2))),(( 0< z<+1)) :} Aplication numerique: c=sin 18=0,309016994 z=(1/2)⇒ sin (x+18)=sin (30) x+18=30 [x=12]$
	$△ A D C ∡ C D A = x + 18; A C = b C D =$ $\frac{\sin 18}{b} = \frac{\sin (x + 18)}{a} \Rightarrow \frac{a}{b} = \frac{\sin (x + 18)}{\sin 18} (1)$ $△ A B C B D = A C = a B C = a + b$ $\frac{\sin (x + 18)}{a + b} = \frac{\sin 18}{a}$ $a [\sin (x + 18) - \sin 18] = b \sin 18$ $\frac{a}{b} = \frac{\sin 18}{\sin (x + 18) - \sin 18} (2)$ $(1) = (2) \Rightarrow$ $\frac{\sin (x + 18)}{\sin 18} = \frac{\sin 18}{\sin (x + 18) - \sin 18}$ $p o s o n s z = \sin (x + 18) \sin 18 = c$ $\frac{z}{c} = \frac{c}{z - c} \Rightarrow [z^{2} - c z - c^{2} = 0]$ $△ = c^{2} + 4 c {\begin{cases} z = \frac{c \pm \sqrt{c^{2} + 4 c^{2}}}{2} = \frac{c (1 + \sqrt{5})}{2} \\ 0 < z < + 1 \end{cases}$ $A p l i c a t i o n n u m e r i q u e :$ $c = \sin 18 = 0, 309016994$ $z = \frac{1}{2} \Rightarrow \sin (x + 18) = \sin (30)$ $x + 18 = 30$ $[x = 12]$
	Commented by a.lgnaoui last updated on 25/Feb/23

	Commented by Shrinava last updated on 25/Feb/23

	$cool dear professor thank you$
	Commented by a.lgnaoui last updated on 26/Feb/23

	$t h a n k y o u v e r y m u c h$
	Answered by infinityaction last updated on 26/Feb/23

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