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Question Number 209913 by mr W last updated on 25/Jul/24

Commented by mr W last updated on 25/Jul/24

$A B i s d i a m e t e r, C D i s t a n g e n t .$ $A E = E D, A B / / C D$ $f i n d a n g l e x = ?$
	Answered by mahdipoor last updated on 26/Jul/24

	$d e f i n e :$ ${E H}_{1} (⊥ B A), {E H}_{2} (⊥ C D), {A H}_{3} (⊥ C D)$ ${A H}_{3} D \sim {E H}_{2} D \Rightarrow \frac{A D}{E D} = \frac{{A H}_{3}}{{E H}_{2}} \Rightarrow \frac{2}{1} = \frac{R}{{E H}_{3}} \Rightarrow {E H}_{3} = \frac{R}{2}$ ${E H}_{3} + {E H}_{1} = R \Rightarrow {E H}_{1} = \frac{R}{2}$ $S_{A O E} = \frac{{E H}_{1} \times A O}{2} = \frac{A O \times O E \times s i n (A O E)}{2}$ $\Rightarrow \frac{1}{2} = s i n (180 - 2 x) \Rightarrow x = 15$
	Answered by a.lgnaoui last updated on 26/Jul/24
	$((sin x)/(AC))=((sin (45−x))/(CD)) sin x=(R/(AD)) AC=R(√2) (R/(AD×R(√2)))=((√2)/(2CD))(cos x−sin x) (1/(AD)) =(((cos x−sin x))/(CD)) CD^2 =DE×AD=((AD^2 )/2)⇒ CD=((AD(√2))/2) Donc (1/(AD))=(((cos x−sin x)(√2))/(AD)) soit ((√2)/2)=cos x−sin x cos x(1−tan x)−((√2)/2)=0 tan x=t ((1−t)/( (√(1+t^2 ))))=(1/( (√2))) (((1−t)^2 )/(1+t^2 ))=(1/2) 2−4t+2t^2 =1+t^2 t^2 −4t+3=0 t=((2±(√3))/1) soit { ((x=75°)),((x=15°)) :} or x<45° ⇒ x=15° est solution$
	$\frac{\sin x}{AC} = \frac{\sin (45 - x)}{CD}$ $\sin x = \frac{R}{AD} AC = R \sqrt{2}$ $\frac{R}{AD \times R \sqrt{2}} = \frac{\sqrt{2}}{2 CD} (\cos x - \sin x)$ $\frac{1}{AD} = \frac{(\cos x - \sin x)}{CD}$ ${CD}^{2} = DE \times AD = \frac{{AD}^{2}}{2} \Rightarrow CD = \frac{AD \sqrt{2}}{2}$ $Donc$ $\frac{1}{AD} = \frac{(\cos x - \sin x) \sqrt{2}}{AD}$ $soit \frac{\sqrt{2}}{2} = \cos x - \sin x$ $\cos x (1 - \tan x) - \frac{\sqrt{2}}{2} = 0$ $\tan x = t$ $\frac{1 - t}{\sqrt{1 + t^{2}}} = \frac{1}{\sqrt{2}} \frac{{(1 - t)}^{2}}{1 + t^{2}} = \frac{1}{2}$ $2 - 4 t + 2 t^{2} = 1 + t^{2}$ $t^{2} - 4 t + 3 = 0$ $t = \frac{2 \pm \sqrt{3}}{1} soit {\begin{cases} x = 75 ° \\ x = 15 ° \end{cases}$ $or x < 45 ° \Rightarrow x = 15 ° est solution$
	Answered by mr W last updated on 26/Jul/24

	Commented by mr W last updated on 26/Jul/24

	$\frac{A B}{A E} = \frac{A D}{F D} \Rightarrow \frac{2 R}{A E} = \frac{A D}{R + C D}$ $A D \times E D = A D \times A E = 2 R (R + C D) = {C D}^{2}$ $\Rightarrow {C D}^{2} - 2 R \times C D = 2 R^{2}$ $\Rightarrow C D = (\sqrt{3} + 1) R$ $\tan x = \frac{R}{R + (\sqrt{3} + 1) R} = 2 - \sqrt{3}$ $\Rightarrow x = 15 °$
	Answered by A5T last updated on 26/Jul/24

	Commented by A5T last updated on 26/Jul/24

	$D E \times D A = 2 {D E}^{2} = {C D}^{2} \Rightarrow C D = D E \sqrt{2}$ $\frac{s i n 135}{2 D E} = \frac{s i n (45 - x)}{C D} \Rightarrow s i n (45 - x) = \frac{C D s i n 135 °}{2 D E}$ $\Rightarrow s i n (45 - x) = \frac{1}{2} \Rightarrow 45 - x = 30 ° \Rightarrow x = 15 °$
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