Question-172558

Question Number 172558 by bagjagugum123 last updated on 28/Jun/22

Answered by mr W last updated on 28/Jun/22

Commented by mr W last updated on 28/Jun/22

((AC)/(sin (36−α)))=(1/(sin (36+α)))=((BC)/(sin 72)) ⇒AC=((sin (36−α))/(sin (36+α))) ⇒BC=((sin 72)/(sin (36+α))) ((AD)/(sin (36+α)))=((CD)/(sin 36))=((AC)/(sin α)) AC=((sin α CD)/(sin 36))=((sin α sin 72)/(sin 36 sin (36+α))) ((sin α sin 72)/(sin 36 sin (36+α)))=((sin (36−α))/(sin (36+α))) 2 sin α cos 36=sin 36 cos α−cos 36 sin α ⇒tan α=((tan 36)/3) AD=((sin (36+α))/(sin α))×((sin (36−α))/(sin (36+α))) AD=((sin (36−α))/(sin α))=((sin 36)/(tan α))−cos 36 AD=3 cos 36−cos 36=2 cos 36=(((√5)+1)/2) ⇒?=AD=(((√5)+1)/2)=φ ✓

$$\frac{{AC}}{\mathrm{sin}\:\left(\mathrm{36}−\alpha\right)}=\frac{\mathrm{1}}{\mathrm{sin}\:\left(\mathrm{36}+\alpha\right)}=\frac{{BC}}{\mathrm{sin}\:\mathrm{72}} \\ $$$$\Rightarrow{AC}=\frac{\mathrm{sin}\:\left(\mathrm{36}−\alpha\right)}{\mathrm{sin}\:\left(\mathrm{36}+\alpha\right)} \\ $$$$\Rightarrow{BC}=\frac{\mathrm{sin}\:\mathrm{72}}{\mathrm{sin}\:\left(\mathrm{36}+\alpha\right)} \\ $$$$\frac{{AD}}{\mathrm{sin}\:\left(\mathrm{36}+\alpha\right)}=\frac{{CD}}{\mathrm{sin}\:\mathrm{36}}=\frac{{AC}}{\mathrm{sin}\:\alpha} \\ $$$${AC}=\frac{\mathrm{sin}\:\alpha\:{CD}}{\mathrm{sin}\:\mathrm{36}}=\frac{\mathrm{sin}\:\alpha\:\mathrm{sin}\:\mathrm{72}}{\mathrm{sin}\:\mathrm{36}\:\mathrm{sin}\:\left(\mathrm{36}+\alpha\right)} \\ $$$$ \\ $$$$\frac{\mathrm{sin}\:\alpha\:\mathrm{sin}\:\mathrm{72}}{\mathrm{sin}\:\mathrm{36}\:\mathrm{sin}\:\left(\mathrm{36}+\alpha\right)}=\frac{\mathrm{sin}\:\left(\mathrm{36}−\alpha\right)}{\mathrm{sin}\:\left(\mathrm{36}+\alpha\right)} \\ $$$$\mathrm{2}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\mathrm{36}=\mathrm{sin}\:\mathrm{36}\:\mathrm{cos}\:\alpha−\mathrm{cos}\:\mathrm{36}\:\mathrm{sin}\:\alpha \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{tan}\:\mathrm{36}}{\mathrm{3}} \\ $$$$ \\ $$$${AD}=\frac{\mathrm{sin}\:\left(\mathrm{36}+\alpha\right)}{\mathrm{sin}\:\alpha}×\frac{\mathrm{sin}\:\left(\mathrm{36}−\alpha\right)}{\mathrm{sin}\:\left(\mathrm{36}+\alpha\right)} \\ $$$${AD}=\frac{\mathrm{sin}\:\left(\mathrm{36}−\alpha\right)}{\mathrm{sin}\:\alpha}=\frac{\mathrm{sin}\:\mathrm{36}}{\mathrm{tan}\:\alpha}−\mathrm{cos}\:\mathrm{36} \\ $$$${AD}=\mathrm{3}\:\mathrm{cos}\:\mathrm{36}−\mathrm{cos}\:\mathrm{36}=\mathrm{2}\:\mathrm{cos}\:\mathrm{36}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow?={AD}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}=\phi\:\checkmark \\ $$

Commented by bagjagugum123 last updated on 29/Jun/22

$${Nice}\:{Solution}\:{Sir}\: \\ $$$${Thanks} \\ $$$$ \\ $$

Commented by Tawa11 last updated on 30/Jun/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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