Question Number 214629 by ajfour last updated on 14/Dec/24
Commented by ajfour last updated on 14/Dec/24
$${say}\:{for}\:{a}=\mathrm{3},\:{b}=\mathrm{2}. \\ $$
Answered by mr W last updated on 14/Dec/24
Commented by mr W last updated on 14/Dec/24
![say (b/a)=μ, (r/a)=λ tan 2α=(b/a)=μ ⇒tan α=(((√(1+μ^2 ))−1)/μ) 2γ=2β+(π/2)−2α ⇒γ=(π/4)+β−α 2β+2δ=(π/2) ⇒δ=(π/4)−β ((CD)/(sin 2α))=(a/(sin 2γ)) ⇒CD=((a sin 2α)/(sin 2γ)) =((a sin 2α)/(cos (2β−2α))) =(a/(((cos 2β)/(tan 2α))+sin 2β)) =(a/(((1−tan^2 β)/(μ(1+tan^2 β)))+((2 tan β)/(1+tan^2 β)))) =((a(1+tan^2 β))/(((1−tan^2 β)/μ)+2 tan β)) CD=(r/(tan γ))+(r/(tan δ)) ⇒((CD)/r)=((1−tan (β−α))/(1+tan (β−α)))+((1+tan β)/(1−tan β)) CD=(r/(tan β))−(r/(tan γ)) ⇒((CD)/r)=(1/(tan β))−((1−tan (β−α))/(1+tan (β−α))) ((1−tan (β−α))/(1+tan (β−α)))+((1+tan β)/(1−tan β))=(1/(tan β))−((1−tan (β−α))/(1+tan (β−α))) 2×((1−tan (β−α))/(1+tan (β−α)))=(1/(tan β))−((1+tan β)/(1−tan β)) (4/(1+tan (β−α)))−2=1+(1/(tan β))−(2/(1−tan β)) (4/(1+((tan β−tan α)/(1+tan α tan β))))−3=(1/(tan β))−(2/(1−tan β)) ((4(1+tan α tan β))/(1−tan α+(1+tan α)tan β))−(1/(tan β))+(2/(1−tan β))−3=0 ((a(1+tan^2 β))/(((1−tan^2 β)/μ)+2 tan β))=r[(1/(tan β))−((1−tan (β−α))/(1+tan (β−α)))] ((a(1+tan^2 β))/(((1−tan^2 β)/μ)+2 tan β))=r[1+(1/(tan β))−((2(1+tan α tan β))/(1−tan α+(1+tan α)tan β))] ⇒λ=((1+tan^2 β)/((((1−tan^2 β)/μ)+2 tan β)[1+(1/(tan β))−((2(1+tan α tan β))/(1−tan α+(1+tan α)tan β))])) example: a=3, b=2 ⇒λ=0.2 ⇒r=0.6](https://www.tinkutara.com/question/Q214634.png)
$${say}\:\frac{{b}}{{a}}=\mu,\:\frac{{r}}{{a}}=\lambda \\ $$$$\mathrm{tan}\:\mathrm{2}\alpha=\frac{{b}}{{a}}=\mu \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }−\mathrm{1}}{\mu} \\ $$$$\mathrm{2}\gamma=\mathrm{2}\beta+\frac{\pi}{\mathrm{2}}−\mathrm{2}\alpha \\ $$$$\Rightarrow\gamma=\frac{\pi}{\mathrm{4}}+\beta−\alpha \\ $$$$\mathrm{2}\beta+\mathrm{2}\delta=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\delta=\frac{\pi}{\mathrm{4}}−\beta \\ $$$$\frac{{CD}}{\mathrm{sin}\:\mathrm{2}\alpha}=\frac{{a}}{\mathrm{sin}\:\mathrm{2}\gamma} \\ $$$$\Rightarrow{CD}=\frac{{a}\:\mathrm{sin}\:\mathrm{2}\alpha}{\mathrm{sin}\:\mathrm{2}\gamma} \\ $$$$\:\:\:\:\:=\frac{{a}\:\mathrm{sin}\:\mathrm{2}\alpha}{\mathrm{cos}\:\left(\mathrm{2}\beta−\mathrm{2}\alpha\right)} \\ $$$$\:\:\:\:\:=\frac{{a}}{\frac{\mathrm{cos}\:\mathrm{2}\beta}{\mathrm{tan}\:\mathrm{2}\alpha}+\mathrm{sin}\:\mathrm{2}\beta} \\ $$$$\:\:\:\:\:=\frac{{a}}{\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\beta}{\mu\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\beta\right)}+\frac{\mathrm{2}\:\mathrm{tan}\:\beta}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\beta}} \\ $$$$\:\:\:\:\:=\frac{{a}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\beta\right)}{\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\beta}{\mu}+\mathrm{2}\:\mathrm{tan}\:\beta} \\ $$$${CD}=\frac{{r}}{\mathrm{tan}\:\gamma}+\frac{{r}}{\mathrm{tan}\:\delta} \\ $$$$\Rightarrow\frac{{CD}}{{r}}=\frac{\mathrm{1}−\mathrm{tan}\:\left(\beta−\alpha\right)}{\mathrm{1}+\mathrm{tan}\:\left(\beta−\alpha\right)}+\frac{\mathrm{1}+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\beta} \\ $$$${CD}=\frac{{r}}{\mathrm{tan}\:\beta}−\frac{{r}}{\mathrm{tan}\:\gamma} \\ $$$$\Rightarrow\frac{{CD}}{{r}}=\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\frac{\mathrm{1}−\mathrm{tan}\:\left(\beta−\alpha\right)}{\mathrm{1}+\mathrm{tan}\:\left(\beta−\alpha\right)} \\ $$$$\frac{\mathrm{1}−\mathrm{tan}\:\left(\beta−\alpha\right)}{\mathrm{1}+\mathrm{tan}\:\left(\beta−\alpha\right)}+\frac{\mathrm{1}+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\beta}=\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\frac{\mathrm{1}−\mathrm{tan}\:\left(\beta−\alpha\right)}{\mathrm{1}+\mathrm{tan}\:\left(\beta−\alpha\right)} \\ $$$$\mathrm{2}×\frac{\mathrm{1}−\mathrm{tan}\:\left(\beta−\alpha\right)}{\mathrm{1}+\mathrm{tan}\:\left(\beta−\alpha\right)}=\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\frac{\mathrm{1}+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\beta} \\ $$$$\frac{\mathrm{4}}{\mathrm{1}+\mathrm{tan}\:\left(\beta−\alpha\right)}−\mathrm{2}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\frac{\mathrm{2}}{\mathrm{1}−\mathrm{tan}\:\beta} \\ $$$$\frac{\mathrm{4}}{\mathrm{1}+\frac{\mathrm{tan}\:\beta−\mathrm{tan}\:\alpha}{\mathrm{1}+\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta}}−\mathrm{3}=\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\frac{\mathrm{2}}{\mathrm{1}−\mathrm{tan}\:\beta} \\ $$$$\frac{\mathrm{4}\left(\mathrm{1}+\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\right)}{\mathrm{1}−\mathrm{tan}\:\alpha+\left(\mathrm{1}+\mathrm{tan}\:\alpha\right)\mathrm{tan}\:\beta}−\frac{\mathrm{1}}{\mathrm{tan}\:\beta}+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{tan}\:\beta}−\mathrm{3}=\mathrm{0} \\ $$$$\frac{{a}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\beta\right)}{\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\beta}{\mu}+\mathrm{2}\:\mathrm{tan}\:\beta}={r}\left[\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\frac{\mathrm{1}−\mathrm{tan}\:\left(\beta−\alpha\right)}{\mathrm{1}+\mathrm{tan}\:\left(\beta−\alpha\right)}\right] \\ $$$$\frac{{a}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\beta\right)}{\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\beta}{\mu}+\mathrm{2}\:\mathrm{tan}\:\beta}={r}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\right)}{\mathrm{1}−\mathrm{tan}\:\alpha+\left(\mathrm{1}+\mathrm{tan}\:\alpha\right)\mathrm{tan}\:\beta}\right] \\ $$$$\Rightarrow\lambda=\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\beta}{\left(\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\beta}{\mu}+\mathrm{2}\:\mathrm{tan}\:\beta\right)\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\right)}{\mathrm{1}−\mathrm{tan}\:\alpha+\left(\mathrm{1}+\mathrm{tan}\:\alpha\right)\mathrm{tan}\:\beta}\right]} \\ $$$$ \\ $$$${example}:\: \\ $$$${a}=\mathrm{3},\:{b}=\mathrm{2}\:\Rightarrow\lambda=\mathrm{0}.\mathrm{2}\:\Rightarrow{r}=\mathrm{0}.\mathrm{6} \\ $$
Commented by mr W last updated on 14/Dec/24
Commented by mr W last updated on 14/Dec/24
Commented by mr W last updated on 14/Dec/24
