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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
AB is diameter, CD is tangent. AE=ED, AB//CD find angle x=?
$${AB}\:{is}\:{diameter},\:{CD}\:{is}\:{tangent}.\: \\ $$$${AE}={ED},\:{AB}//{CD} \\ $$$${find}\:{angle}\:{x}=? \\ $$
Answered by mahdipoor last updated on 26/Jul/24
define: EH_1 (⊥BA ) , EH_(2 ) (⊥CD) , AH_3 (⊥CD) AH_3 D∼EH_2 D⇒((AD)/(ED))=((AH_3 )/(EH_2 ))⇒(2/1)=(R/(EH_3 ))⇒EH_3 =(R/2) EH_3 +EH_1 =R⇒EH_1 =(R/2) S_(AOE) =((EH_1 ×AO)/2)=((AO×OE×sin(AOE))/2) ⇒(1/2)=sin(180−2x)⇒x=15
$${define}: \\ $$$${EH}_{\mathrm{1}} \:\left(\bot{BA}\:\right)\:,\:\:{EH}_{\mathrm{2}\:\:} \left(\bot{CD}\right)\:,\:\:{AH}_{\mathrm{3}} \:\:\left(\bot{CD}\right) \\ $$$${AH}_{\mathrm{3}} {D}\sim{EH}_{\mathrm{2}} {D}\Rightarrow\frac{{AD}}{{ED}}=\frac{{AH}_{\mathrm{3}} }{{EH}_{\mathrm{2}} }\Rightarrow\frac{\mathrm{2}}{\mathrm{1}}=\frac{{R}}{{EH}_{\mathrm{3}} }\Rightarrow{EH}_{\mathrm{3}} =\frac{{R}}{\mathrm{2}} \\ $$$${EH}_{\mathrm{3}} +{EH}_{\mathrm{1}} ={R}\Rightarrow{EH}_{\mathrm{1}} =\frac{{R}}{\mathrm{2}} \\ $$$${S}_{{AOE}} =\frac{{EH}_{\mathrm{1}} ×{AO}}{\mathrm{2}}=\frac{{AO}×{OE}×{sin}\left({AOE}\right)}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}={sin}\left(\mathrm{180}−\mathrm{2}{x}\right)\Rightarrow{x}=\mathrm{15} \\ $$
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
$$\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{AC}}=\frac{\mathrm{sin}\:\left(\mathrm{45}−\mathrm{x}\right)}{\mathrm{CD}} \\ $$$$\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{R}}{\mathrm{AD}}\:\:\:\mathrm{AC}=\mathrm{R}\sqrt{\mathrm{2}} \\ $$$$\frac{\mathrm{R}}{\mathrm{AD}×\mathrm{R}\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2CD}}\left(\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\right) \\ $$$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{AD}}\:\:\:=\frac{\left(\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\right)}{\mathrm{CD}} \\ $$$$\mathrm{CD}^{\mathrm{2}} =\mathrm{DE}×\mathrm{AD}=\frac{\mathrm{AD}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow\:\mathrm{CD}=\frac{\mathrm{AD}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{Donc}} \\ $$$$\:\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{AD}}}=\frac{\left(\mathrm{cos}\:\boldsymbol{\mathrm{x}}−\mathrm{sin}\:\boldsymbol{\mathrm{x}}\right)\sqrt{\mathrm{2}}}{\boldsymbol{\mathrm{AD}}} \\ $$$$\boldsymbol{\mathrm{soit}}\:\:\:\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{cos}\:\boldsymbol{\mathrm{x}}−\mathrm{sin}\:\boldsymbol{\mathrm{x}} \\ $$$$\mathrm{cos}\:\boldsymbol{\mathrm{x}}\left(\mathrm{1}−\mathrm{tan}\:\mathrm{x}\right)−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{0} \\ $$$$\:\:\mathrm{tan}\:\mathrm{x}=\boldsymbol{\mathrm{t}} \\ $$$$\frac{\mathrm{1}−\boldsymbol{\mathrm{t}}}{\:\sqrt{\mathrm{1}+\boldsymbol{\mathrm{t}}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\:\:\frac{\left(\mathrm{1}−\boldsymbol{\mathrm{t}}\right)^{\mathrm{2}} }{\mathrm{1}+\boldsymbol{\mathrm{t}}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\mathrm{2}−\mathrm{4}\boldsymbol{\mathrm{t}}+\mathrm{2}\boldsymbol{\mathrm{t}}^{\mathrm{2}} =\mathrm{1}+\boldsymbol{\mathrm{t}}^{\mathrm{2}} \\ $$$$\:\:\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{t}}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\boldsymbol{\mathrm{t}}=\frac{\mathrm{2}\pm\sqrt{\mathrm{3}}}{\mathrm{1}}\:\:\:\mathrm{soit}\:\:\:\:\begin{cases}{\mathrm{x}=\mathrm{75}°}\\{\mathrm{x}=\mathrm{15}°}\end{cases} \\ $$$$\boldsymbol{\mathrm{or}}\:\:\boldsymbol{\mathrm{x}}<\mathrm{45}°\:\:\:\Rightarrow\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{15}°\:\mathrm{est}\:\mathrm{solution} \\ $$$$ \\ $$
Answered by mr W last updated on 26/Jul/24
Commented by mr W last updated on 26/Jul/24
((AB)/(AE))=((AD)/(FD)) ⇒((2R)/(AE))=((AD)/(R+CD)) AD×ED=AD×AE=2R(R+CD)=CD^2 ⇒CD^2 −2R×CD=2R^2 ⇒CD=((√3)+1)R tan x=(R/(R+((√3)+1)R))=2−(√3) ⇒x=15°
$$\frac{{AB}}{{AE}}=\frac{{AD}}{{FD}}\:\Rightarrow\frac{\mathrm{2}{R}}{{AE}}=\frac{{AD}}{{R}+{CD}} \\ $$$${AD}×{ED}={AD}×{AE}=\mathrm{2}{R}\left({R}+{CD}\right)={CD}^{\mathrm{2}} \\ $$$$\Rightarrow{CD}^{\mathrm{2}} −\mathrm{2}{R}×{CD}=\mathrm{2}{R}^{\mathrm{2}} \\ $$$$\Rightarrow{CD}=\left(\sqrt{\mathrm{3}}+\mathrm{1}\right){R} \\ $$$$\mathrm{tan}\:{x}=\frac{{R}}{{R}+\left(\sqrt{\mathrm{3}}+\mathrm{1}\right){R}}=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{x}=\mathrm{15}° \\ $$
Answered by A5T last updated on 26/Jul/24
Commented by A5T last updated on 26/Jul/24
DE×DA=2DE^2 =CD^2 ⇒CD=DE(√2) ((sin135)/(2DE))=((sin(45−x))/(CD))⇒sin(45−x)=((CDsin135°)/(2DE)) ⇒sin(45−x)=(1/2)⇒45−x=30°⇒x=15°
$${DE}×{DA}=\mathrm{2}{DE}^{\mathrm{2}} ={CD}^{\mathrm{2}} \Rightarrow{CD}={DE}\sqrt{\mathrm{2}} \\ $$$$\frac{{sin}\mathrm{135}}{\mathrm{2}{DE}}=\frac{{sin}\left(\mathrm{45}−{x}\right)}{{CD}}\Rightarrow{sin}\left(\mathrm{45}−{x}\right)=\frac{{CDsin}\mathrm{135}°}{\mathrm{2}{DE}} \\ $$$$\Rightarrow{sin}\left(\mathrm{45}−{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{45}−{x}=\mathrm{30}°\Rightarrow{x}=\mathrm{15}° \\ $$

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