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Question Number 171765 by cherokeesay last updated on 20/Jun/22

Answered by som(math1967) last updated on 21/Jun/22

let OB=OA=OM=r ∴AC=(√(r^2 +(r^2 /4)))=((r(√5))/2) △AOC∼△OLC ∴((LC)/(OC))=((r/2)/((r(√5))/2)) ⇒LC=(r/(2(√5))) ∴AL=((r(√5))/2) −(r/(2(√5)))=((2r)/( (√5))) same way ((OL)/(OC))=(r/((r(√5))/2)) OL=(2/( (√5)))×(r/2)=(r/( (√5))) LM=r−(r/( (√5)))=((((√5)−1)r)/( (√5))) ∴ tanβ=((AL)/(LM))=(((2r)/( (√5)))/((((√5)−1)r)/( (√5))))=(2/( (√5)−1)) ∴ tanβ=((2((√5)+1))/(5−1))=(((√5)+1)/2)=ϕ golden ratio

$${let}\:{OB}={OA}={OM}={r} \\ $$$$\:\therefore{AC}=\sqrt{{r}^{\mathrm{2}} +\frac{{r}^{\mathrm{2}} }{\mathrm{4}}}=\frac{{r}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\bigtriangleup{AOC}\sim\bigtriangleup{OLC} \\ $$$$\:\therefore\frac{{LC}}{{OC}}=\frac{\frac{{r}}{\mathrm{2}}}{\frac{{r}\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$$$\Rightarrow{LC}=\frac{{r}}{\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\therefore{AL}=\frac{{r}\sqrt{\mathrm{5}}}{\mathrm{2}}\:−\frac{{r}}{\mathrm{2}\sqrt{\mathrm{5}}}=\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{5}}} \\ $$$${same}\:{way} \\ $$$$\frac{{OL}}{{OC}}=\frac{{r}}{\frac{{r}\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$$${OL}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}×\frac{{r}}{\mathrm{2}}=\frac{{r}}{\:\sqrt{\mathrm{5}}} \\ $$$${LM}={r}−\frac{{r}}{\:\sqrt{\mathrm{5}}}=\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){r}}{\:\sqrt{\mathrm{5}}} \\ $$$$\:\therefore\:{tan}\beta=\frac{{AL}}{{LM}}=\frac{\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{5}}}}{\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){r}}{\:\sqrt{\mathrm{5}}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}−\mathrm{1}} \\ $$$$\therefore\:{tan}\beta=\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}{\mathrm{5}−\mathrm{1}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}=\varphi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{golden}}\:\boldsymbol{{ratio}} \\ $$$$ \\ $$

Commented by som(math1967) last updated on 21/Jun/22

Commented by Tawa11 last updated on 21/Jun/22

Great sir

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

Commented by cherokeesay last updated on 21/Jun/22

Nice ....thank you sir.

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

Answered by mr W last updated on 21/Jun/22

Commented by mr W last updated on 21/Jun/22

tan α=(2/1)=((2 tan (α/2))/(1−tan^2 (α/2))) ⇒((1/(tan (α/2))))^2 −(1/(tan (α/2)))−1=0 ⇒(1/(tan (α/2)))=((1+(√5))/2)=ϕ α+2β=π ⇒β=(π/2)−(α/2) ⇒tan β=(1/(tan (α/2)))=((1+(√5))/2)=ϕ ✓

$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}}{\mathrm{1}}=\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\varphi \\ $$$$ \\ $$$$\alpha+\mathrm{2}\beta=\pi\:\Rightarrow\beta=\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}\:\beta=\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\varphi\:\checkmark \\ $$

Commented by Tawa11 last updated on 21/Jun/22

Great sir

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

Commented by cherokeesay last updated on 21/Jun/22

thank you mister.

$${thank}\:{you}\:{mister}. \\ $$

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