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Question-27429

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Question Number 27429 by Tinkutara last updated on 06/Jan/18
Answered by mrW1 last updated on 07/Jan/18
let′s see point 2 as example, ((v_2 cos θ_2 )/(v_1 cos θ_1 ))=e ⇒ v_2 cos θ_2 =e v_1 cos θ_1 ...(i) mv_2 sin θ_2 =mv_1 sin θ_1 ⇒ v_2 sin θ_2 =v_1 sin θ_1 ...(ii) (ii)/(i): ⇒ tan θ_2 =(1/e) tan θ_1 similarly at point 1: ⇒tan θ_1 =(1/e) tan θ ⇒tan θ_2 =(1/e) tan θ_1 =(1/e^2 ) tan θ 2(θ_2 +θ_1 +θ)=π θ_2 +θ_1 =(π/2)−θ tan (θ_2 +θ_1 )=tan ((π/2)−θ) ((tan θ_2 +tan θ_1 )/(1−tan θ_1 tan θ_2 ))=(1/(tan θ)) ((((tan θ)/e^2 )+((tan θ)/e))/(1−((tan^2 θ)/e^3 )))=(1/(tan θ)) ((e(e+1)tan θ)/(e^3 −tan^2 θ))=(1/(tan θ)) e(e+1)tan^2 θ=e^3 −tan^2 θ (e^2 +e+1)tan^2 θ=e^3 ⇒ tan^2 θ=(e^3 /(e^2 +e+1)) ⇒Answer (4)
$${let}'{s}\:{see}\:{point}\:\mathrm{2}\:{as}\:{example}, \\ $$$$\frac{{v}_{\mathrm{2}} \:\mathrm{cos}\:\theta_{\mathrm{2}} }{{v}_{\mathrm{1}} \:\mathrm{cos}\:\theta_{\mathrm{1}} }={e} \\ $$$$\Rightarrow\:{v}_{\mathrm{2}} \:\mathrm{cos}\:\theta_{\mathrm{2}} ={e}\:{v}_{\mathrm{1}} \:\mathrm{cos}\:\theta_{\mathrm{1}} \:\:\:…\left({i}\right) \\ $$$${mv}_{\mathrm{2}} \:\mathrm{sin}\:\theta_{\mathrm{2}} ={mv}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{1}} \\ $$$$\Rightarrow\:{v}_{\mathrm{2}} \:\mathrm{sin}\:\theta_{\mathrm{2}} ={v}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{1}} \:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)/\left({i}\right): \\ $$$$\Rightarrow\:\mathrm{tan}\:\theta_{\mathrm{2}} =\frac{\mathrm{1}}{{e}}\:\mathrm{tan}\:\theta_{\mathrm{1}} \\ $$$$ \\ $$$${similarly}\:{at}\:{point}\:\mathrm{1}: \\ $$$$\Rightarrow\mathrm{tan}\:\theta_{\mathrm{1}} =\frac{\mathrm{1}}{{e}}\:\mathrm{tan}\:\theta \\ $$$$ \\ $$$$\Rightarrow\mathrm{tan}\:\theta_{\mathrm{2}} =\frac{\mathrm{1}}{{e}}\:\mathrm{tan}\:\theta_{\mathrm{1}} =\frac{\mathrm{1}}{{e}^{\mathrm{2}} }\:\mathrm{tan}\:\theta \\ $$$$ \\ $$$$\mathrm{2}\left(\theta_{\mathrm{2}} +\theta_{\mathrm{1}} +\theta\right)=\pi \\ $$$$\theta_{\mathrm{2}} +\theta_{\mathrm{1}} =\frac{\pi}{\mathrm{2}}−\theta \\ $$$$\mathrm{tan}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{1}} \right)=\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right) \\ $$$$\frac{\mathrm{tan}\:\theta_{\mathrm{2}} +\mathrm{tan}\:\theta_{\mathrm{1}} }{\mathrm{1}−\mathrm{tan}\:\theta_{\mathrm{1}} \:\mathrm{tan}\:\theta_{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{tan}\:\theta} \\ $$$$\frac{\frac{\mathrm{tan}\:\theta}{{e}^{\mathrm{2}} }+\frac{\mathrm{tan}\:\theta}{{e}}}{\mathrm{1}−\frac{\mathrm{tan}^{\mathrm{2}} \:\theta}{{e}^{\mathrm{3}} }}=\frac{\mathrm{1}}{\mathrm{tan}\:\theta} \\ $$$$\frac{{e}\left({e}+\mathrm{1}\right)\mathrm{tan}\:\theta}{{e}^{\mathrm{3}} −\mathrm{tan}^{\mathrm{2}} \:\theta}=\frac{\mathrm{1}}{\mathrm{tan}\:\theta} \\ $$$${e}\left({e}+\mathrm{1}\right)\mathrm{tan}^{\mathrm{2}} \:\theta={e}^{\mathrm{3}} −\mathrm{tan}^{\mathrm{2}} \:\theta \\ $$$$\left({e}^{\mathrm{2}} +{e}+\mathrm{1}\right)\mathrm{tan}^{\mathrm{2}} \:\theta={e}^{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{tan}^{\mathrm{2}} \:\theta=\frac{{e}^{\mathrm{3}} }{{e}^{\mathrm{2}} +{e}+\mathrm{1}} \\ $$$$ \\ $$$$\Rightarrow{Answer}\:\left(\mathrm{4}\right) \\ $$
Commented by mrW1 last updated on 07/Jan/18
Commented by Tinkutara last updated on 08/Jan/18
Thank you very much Sir! I got the answer.

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