Question-88525

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Question Number 88525 by jagoll last updated on 11/Apr/20

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{i}\:\mathrm{got}\:\mathrm{20}^{\mathrm{o}} \\ $$

Commented by mr W last updated on 11/Apr/20

$${x}=\mathrm{80}° \\ $$

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{why}? \\ $$

Commented by mahdi last updated on 11/Apr/20

in ΔAOC:((sinx)/(AO))=((sin20)/(OC)) in ΔOBC:((sin(110−x))/(OB))=((sin10)/(OC)) ⇒((sinx)/(AO))=((sin20)/((OB.sin10)/(sin(110−x))))⇒(OB=OA=r) ((sinx)/(sin(110−x)))=((sin20)/(sin10))=2cos10=((sin80)/(sin30)) ⇒x=80

$$\mathrm{in}\:\Delta\mathrm{AOC}:\frac{\mathrm{sinx}}{\mathrm{AO}}=\frac{\mathrm{sin20}}{\mathrm{OC}} \\ $$$$\mathrm{in}\:\Delta\mathrm{OBC}:\frac{\mathrm{sin}\left(\mathrm{110}−\mathrm{x}\right)}{\mathrm{OB}}=\frac{\mathrm{sin10}}{\mathrm{OC}} \\ $$$$\Rightarrow\frac{\mathrm{sinx}}{\mathrm{AO}}=\frac{\mathrm{sin20}}{\frac{\mathrm{OB}.\mathrm{sin10}}{\mathrm{sin}\left(\mathrm{110}−\mathrm{x}\right)}}\Rightarrow\left(\mathrm{OB}=\mathrm{OA}=\mathrm{r}\right) \\ $$$$\frac{\mathrm{sinx}}{\mathrm{sin}\left(\mathrm{110}−\mathrm{x}\right)}=\frac{\mathrm{sin20}}{\mathrm{sin10}}=\mathrm{2cos10}=\frac{\mathrm{sin80}}{\mathrm{sin30}} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{80} \\ $$

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{sir}.\:\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{110}−\mathrm{x}\:?\:\mathrm{sir} \\ $$

Answered by mr W last updated on 11/Apr/20

Commented by mr W last updated on 11/Apr/20

((AD)/(sin 70))=((CA)/(sin 110))=((CA)/(cos 20)) ((CA)/(sin 30))=((AB)/(sin 70)) AB=2 OA cos 20 AD=((sin 70)/(cos 20))×CA=((sin 70)/(cos 20))×((sin 30)/(sin 70))×AB =((sin 70)/(cos 20))×((sin 30)/(sin 70))×2 cos 20×OA =2 sin 30×OA =OA ⇒ΔOAD=isosceles ⇒x=((180−20)/2)=80°

$$\frac{{AD}}{\mathrm{sin}\:\mathrm{70}}=\frac{{CA}}{\mathrm{sin}\:\mathrm{110}}=\frac{{CA}}{\mathrm{cos}\:\mathrm{20}} \\ $$$$\frac{{CA}}{\mathrm{sin}\:\mathrm{30}}=\frac{{AB}}{\mathrm{sin}\:\mathrm{70}} \\ $$$${AB}=\mathrm{2}\:{OA}\:\mathrm{cos}\:\mathrm{20} \\ $$$$ \\ $$$${AD}=\frac{\mathrm{sin}\:\mathrm{70}}{\mathrm{cos}\:\mathrm{20}}×{CA}=\frac{\mathrm{sin}\:\mathrm{70}}{\mathrm{cos}\:\mathrm{20}}×\frac{\mathrm{sin}\:\mathrm{30}}{\mathrm{sin}\:\mathrm{70}}×{AB} \\ $$$$=\frac{\mathrm{sin}\:\mathrm{70}}{\mathrm{cos}\:\mathrm{20}}×\frac{\mathrm{sin}\:\mathrm{30}}{\mathrm{sin}\:\mathrm{70}}×\mathrm{2}\:\mathrm{cos}\:\mathrm{20}×{OA} \\ $$$$=\mathrm{2}\:\mathrm{sin}\:\mathrm{30}×{OA} \\ $$$$={OA} \\ $$$$\Rightarrow\Delta{OAD}={isosceles} \\ $$$$\Rightarrow{x}=\frac{\mathrm{180}−\mathrm{20}}{\mathrm{2}}=\mathrm{80}° \\ $$

Commented by mr W last updated on 11/Apr/20

$${sorry},\:\angle{ADC}=\mathrm{70}° \\ $$$$\mathrm{110}=\mathrm{70}+\mathrm{40} \\ $$

Commented by jagoll last updated on 11/Apr/20

sir W. if ∠ACD = 70^o ∠ADC = 110^o ∠CAD = 40^0 how it? 70^o +110^0 +40^o > 180^o

$$\mathrm{sir}\:\mathrm{W}.\:\mathrm{if}\:\angle\mathrm{ACD}\:=\:\mathrm{70}^{\mathrm{o}} \\ $$$$\angle\mathrm{ADC}\:=\:\mathrm{110}^{\mathrm{o}} \\ $$$$\angle\mathrm{CAD}\:=\:\mathrm{40}^{\mathrm{0}} \\ $$$$\mathrm{how}\:\mathrm{it}?\:\mathrm{70}^{\mathrm{o}} +\mathrm{110}^{\mathrm{0}} +\mathrm{40}^{\mathrm{o}} \:>\:\mathrm{180}^{\mathrm{o}} \\ $$$$ \\ $$

Commented by jagoll last updated on 11/Apr/20