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Question-202976

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Question Number 202976 by ajfour last updated on 06/Jan/24
Commented by ajfour last updated on 06/Jan/24
If tan α=t , find h.
$${If}\:\mathrm{tan}\:\alpha={t}\:,\:{find}\:{h}. \\ $$
Answered by mr W last updated on 06/Jan/24
Commented by mr W last updated on 06/Jan/24
b^2 =(1−h)(1+h)=1−h^2 ⇒b=(√(1−h^2 )) tan α_1 =(h/b) tan α_2 =((1−h)/b) α=α_1 +α_2 tan α=(((h/b)+((1−h)/b))/(1−((h(1−h))/b^2 )))=(√((1+h)/(1−h)))=t ⇒h=((t^2 −1)/(t^2 +1))
$${b}^{\mathrm{2}} =\left(\mathrm{1}−{h}\right)\left(\mathrm{1}+{h}\right)=\mathrm{1}−{h}^{\mathrm{2}} \\ $$$$\Rightarrow{b}=\sqrt{\mathrm{1}−{h}^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\alpha_{\mathrm{1}} =\frac{{h}}{{b}} \\ $$$$\mathrm{tan}\:\alpha_{\mathrm{2}} =\frac{\mathrm{1}−{h}}{{b}} \\ $$$$\alpha=\alpha_{\mathrm{1}} +\alpha_{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\frac{{h}}{{b}}+\frac{\mathrm{1}−{h}}{{b}}}{\mathrm{1}−\frac{{h}\left(\mathrm{1}−{h}\right)}{{b}^{\mathrm{2}} }}=\sqrt{\frac{\mathrm{1}+{h}}{\mathrm{1}−{h}}}={t} \\ $$$$\Rightarrow{h}=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$
Commented by Frix last updated on 06/Jan/24
Yes. And with t=tan α we have h=−cos 2α =1−2cos^2 α
$$\mathrm{Yes}.\:\mathrm{And}\:\mathrm{with}\:{t}=\mathrm{tan}\:\alpha\:\mathrm{we}\:\mathrm{have} \\ $$$${h}=−\mathrm{cos}\:\mathrm{2}\alpha\:=\mathrm{1}−\mathrm{2cos}^{\mathrm{2}} \:\alpha \\ $$
Commented by ajfour last updated on 06/Jan/24
Thanks i like that first rihtaway b gets related to h.
$${Thanks}\:{i}\:{like}\:{that}\:{first}\:{rihtaway} \\ $$$${b}\:{gets}\:{related}\:{to}\:{h}. \\ $$
Answered by a.lgnaoui last updated on 06/Jan/24
2𝛂+λ=π λ=π−2𝛂 AB^2 =2(1+cos 2𝛂) OA^2 =1 AB^2 −OA^2 =(1−h)^2 −h^2 2(1+cos 2𝛂)−1=1−2h 2cos 2𝛂=−2h cos 2𝛂=−h ⇒ h=((t^2 −1)/(1+t^2 ))
$$\mathrm{2}\boldsymbol{\alpha}+\lambda=\pi\:\:\:\:\:\lambda=\pi−\mathrm{2}\boldsymbol{\alpha} \\ $$$$\boldsymbol{\mathrm{AB}}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\boldsymbol{\alpha}\right)\:\:\:\:\boldsymbol{\mathrm{OA}}^{\mathrm{2}} =\mathrm{1} \\ $$$$\boldsymbol{\mathrm{AB}}^{\mathrm{2}} −\boldsymbol{\mathrm{OA}}^{\mathrm{2}} =\left(\mathrm{1}−\boldsymbol{\mathrm{h}}\right)^{\mathrm{2}} −\boldsymbol{\mathrm{h}}^{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\boldsymbol{\alpha}\right)−\mathrm{1}=\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{h}} \\ $$$$\mathrm{2cos}\:\mathrm{2}\boldsymbol{\alpha}=−\mathrm{2h} \\ $$$$\:\:\:\:\mathrm{cos}\:\mathrm{2}\boldsymbol{\alpha}=−\boldsymbol{\mathrm{h}} \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:\:\:\:\boldsymbol{\mathrm{h}}=\frac{\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+\boldsymbol{\mathrm{t}}^{\mathrm{2}} }\:\: \\ $$
Commented by a.lgnaoui last updated on 06/Jan/24

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