Question Number 63108 by ajfour last updated on 29/Jun/19 | ||
Commented by ajfour last updated on 29/Jun/19 | ||
$${Find}\:{x}_{{A}} \:{in}\:{terms}\:{of}\:{a},{b},{s}\:. \\ $$$$\:\:\:\:\:{s}_{{min}} \leqslant{s}\leqslant{s}_{{max}} \:. \\ $$$${a},{b}\:{are}\:{parameters}\:{of}\:{ellipse} \\ $$$${while}\:{s}\:{is}\:{side}\:{length}\:{of}\:\bigtriangleup_{{equilateral}} . \\ $$ | ||
Answered by mr W last updated on 02/Jul/19 | ||
$${R}=\frac{{s}}{\sqrt{\mathrm{3}}}={radius}\:{of}\:{circumcircle} \\ $$$${M}\left({h},{k}\right)={centroid}\:{and}\:{circumcenter} \\ $$$${let}\:{p}=\frac{{h}}{{a}},\:{q}=\frac{{k}}{{b}},\:\alpha=\frac{{R}}{{a}}=\frac{{s}}{\sqrt{\mathrm{3}}{a}},\:\beta=\frac{{R}}{{b}}=\frac{{s}}{\sqrt{\mathrm{3}}{b}} \\ $$$${x}_{{A}} ={h}+{R}\:\mathrm{cos}\:\theta={a}\left(\frac{{h}}{{a}}+\frac{{R}}{{a}}\:\mathrm{cos}\:\theta\right)={a}\left({p}+\alpha\:\mathrm{cos}\:\theta\right) \\ $$$${y}_{{A}} ={k}+{R}\:\mathrm{sin}\:\theta={b}\left(\frac{{k}}{{b}}+\frac{{R}}{{b}}\:\mathrm{sin}\:\theta\right)={b}\left({q}+\beta\:\mathrm{sin}\:\theta\right) \\ $$$$\frac{{x}_{{A}} ^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}_{{A}} ^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow\left({p}+\alpha\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left({q}+\beta\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{p}^{\mathrm{2}} +\mathrm{2}\alpha{p}\:\mathrm{cos}\:\theta+\alpha^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta+{q}^{\mathrm{2}} +\mathrm{2}\beta{q}\:\mathrm{sin}\:\theta+\beta^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta=\mathrm{1}\:\:\:...\left({i}\right) \\ $$$$ \\ $$$${x}_{{B}} ={a}\left[{p}+\alpha\:\mathrm{cos}\:\left(\theta+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right] \\ $$$${y}_{{B}} ={b}\left[{q}+\beta\:\mathrm{sin}\:\left(\theta+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right] \\ $$$$\Rightarrow\left[{p}+\alpha\:\mathrm{cos}\:\left(\theta+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right]^{\mathrm{2}} +\left[{q}+\beta\:\mathrm{sin}\:\left(\theta+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right]^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\left[{p}−\frac{\alpha}{\mathrm{2}}\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right)\right]^{\mathrm{2}} +\left[{q}−\frac{\beta}{\mathrm{2}}\left(\mathrm{sin}\:\theta−\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)\right]^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{p}^{\mathrm{2}} −\alpha{p}\:\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right)+\frac{\alpha^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +{q}^{\mathrm{2}} −\beta{q}\:\left(\mathrm{sin}\:\theta−\sqrt{\mathrm{3}}\:\mathrm{cos}\:\theta\right)+\frac{\beta^{\mathrm{2}} }{\mathrm{4}}\:\left(\mathrm{sin}\:\theta−\sqrt{\mathrm{3}}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} =\mathrm{1}\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$${x}_{{C}} ={a}\left[{p}+\alpha\:\mathrm{cos}\:\left(\theta−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right] \\ $$$${y}_{{C}} ={b}\left[{q}+\beta\:\mathrm{sin}\:\left(\theta−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right] \\ $$$$\Rightarrow\left[{p}+\alpha\:\mathrm{cos}\:\left(\theta−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right]^{\mathrm{2}} +\left[{q}+\beta\:\mathrm{sin}\:\left(\theta−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right]^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\left[{p}−\frac{\alpha}{\mathrm{2}}\left(\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right)\right]^{\mathrm{2}} +\left[{q}−\frac{\beta}{\mathrm{2}}\left(\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)\right]^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{p}^{\mathrm{2}} −\alpha{p}\:\left(\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right)+\frac{\alpha^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +{q}^{\mathrm{2}} −\beta{q}\:\left(\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)+\frac{\beta^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)^{\mathrm{2}} =\mathrm{1}\:\:\:...\left({iii}\right) \\ $$$$ \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\Rightarrow\alpha\left(\mathrm{3cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right){p}+\frac{\mathrm{1}}{\mathrm{4}}\alpha^{\mathrm{2}} \left(\mathrm{3cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right)\left(\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right)+\beta\left(\mathrm{3sin}\:\theta−\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right){q}+\frac{\mathrm{1}}{\mathrm{4}}\beta^{\mathrm{2}} \left(\mathrm{3}\:\mathrm{sin}\:\theta−\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)\left(\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\alpha\left(\mathrm{3cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right){p}+\mathrm{4}\beta\left(\mathrm{3sin}\:\theta−\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right){q}=−\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right)\left(\mathrm{3cos}\:\mathrm{2}\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{2}\theta\right)\:\:\:\:\:...\left({I}\right) \\ $$$$\left({i}\right)−\left({iii}\right): \\ $$$$\Rightarrow\alpha{p}\left(\mathrm{2cos}\:\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right)+\frac{\mathrm{1}}{\mathrm{4}}\alpha^{\mathrm{2}} \left(\mathrm{3cos}\:\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right)\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right)+\beta{q}\left(\mathrm{3sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)+\frac{\mathrm{1}}{\mathrm{4}}\beta^{\mathrm{2}} \left(\mathrm{3sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)\left(\mathrm{sin}\:\theta−\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\alpha\left(\mathrm{3cos}\:\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right){p}+\mathrm{4}\beta\left(\mathrm{3sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right){q}=−\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right)\left(\mathrm{3cos}\:\mathrm{2}\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{2}\theta\right)\:\:\:\:\:...\left({II}\right) \\ $$ | ||
Commented by mr W last updated on 06/Jul/19 | ||
$$\left(\mathrm{3cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right)\alpha{p}+\left(\mathrm{3sin}\:\theta−\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)\beta{q}=−\frac{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{3cos}\:\mathrm{2}\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{2}\theta\right)\:\:\:\:\:...\left({i}\right) \\ $$$$\left(\mathrm{3cos}\:\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right)\alpha{p}+\left(\mathrm{3sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)\beta{q}=−\frac{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{3cos}\:\mathrm{2}\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{2}\theta\right)\:\:\:\:\:...\left({ii}\right) \\ $$$${D}=\left(\mathrm{3cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right)\left(\mathrm{3sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)−\left(\mathrm{3cos}\:\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right)\left(\mathrm{3sin}\:\theta−\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right) \\ $$$$=\left(\mathrm{6sin}\:\mathrm{2}\theta+\mathrm{3}\sqrt{\mathrm{3}}\right)−\left(\mathrm{6sin}\:\mathrm{2}\theta−\mathrm{3}\sqrt{\mathrm{3}}\right) \\ $$$$=\mathrm{6}\sqrt{\mathrm{3}} \\ $$$$−\frac{\mathrm{4}}{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }\left(\alpha{p}\right){D}=−\left(\mathrm{3sin}\:\theta−\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)\left(\mathrm{3cos}\:\mathrm{2}\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{2}\theta\right)+\left(\mathrm{3sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)\left(\mathrm{3cos}\:\mathrm{2}\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{2}\theta\right) \\ $$$$=\mathrm{6}\sqrt{\mathrm{3}}\left(\mathrm{cos}\:\theta\mathrm{cos}\:\mathrm{2}\theta−\mathrm{sin}\:\theta\mathrm{sin}\:\mathrm{2}\theta\right) \\ $$$$=\mathrm{6}\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{3}\theta \\ $$$$\Rightarrow\alpha{p}=−\frac{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }{\mathrm{4}}\mathrm{cos}\:\mathrm{3}\theta \\ $$$$−\frac{\mathrm{4}}{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }\left(\beta{q}\right){D}=−\left(\mathrm{3cos}\:\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right)\left(\mathrm{3cos}\:\mathrm{2}\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{2}\theta\right)+\left(\mathrm{3cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right)\left(\mathrm{3cos}\:\mathrm{2}\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{2}\theta\right) \\ $$$$=\mathrm{6}\sqrt{\mathrm{3}}\left(\mathrm{sin}\:\theta\mathrm{cos}\:\mathrm{2}\theta+\mathrm{cos}\:\theta\mathrm{sin}\:\mathrm{2}\theta\right) \\ $$$$=\mathrm{6}\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{3}\theta \\ $$$$\Rightarrow\beta{q}=−\frac{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }{\mathrm{4}}\mathrm{sin}\:\mathrm{3}\theta \\ $$$${put}\:{them}\:{into}\: \\ $$$$\left({p}+\alpha\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left({q}+\beta\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${we}\:{get} \\ $$$${b}^{\mathrm{2}} \left(\mathrm{cos}\:\theta+\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} }\mathrm{cos}\:\mathrm{3}\theta\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\mathrm{sin}\:\theta+\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }\mathrm{sin}\:\mathrm{3}\theta\right)^{\mathrm{2}} =\frac{\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{s}^{\mathrm{2}} } \\ $$$${b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta\left[\mathrm{1}+\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} }\left(\mathrm{cos}^{\mathrm{2}} \:\theta−\frac{\mathrm{3}}{\mathrm{4}}\right)\right]^{\mathrm{2}} +{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta\left[\mathrm{1}+\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} }\left(\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{sin}^{\mathrm{2}} \:\theta\right)\right]^{\mathrm{2}} =\frac{\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{s}^{\mathrm{2}} } \\ $$$${b}^{\mathrm{2}} \left(\mathrm{cos}^{\mathrm{2}} \:\theta−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left[\mathrm{1}+\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} }\left(\mathrm{cos}^{\mathrm{2}} \:\theta−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right)\right]^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{cos}^{\mathrm{2}} \:\theta+\frac{\mathrm{1}}{\mathrm{2}}\right)\left[\mathrm{1}+\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} }\left(\mathrm{cos}^{\mathrm{2}} \:\theta−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\right]^{\mathrm{2}} =\frac{\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{s}^{\mathrm{2}} } \\ $$$${let}\:\Phi=\mathrm{cos}^{\mathrm{2}} \:\theta−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} \left(\Phi+\frac{\mathrm{1}}{\mathrm{2}}\right)\left[\mathrm{1}+\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} }\left(\Phi−\frac{\mathrm{1}}{\mathrm{4}}\right)\right]^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\Phi\right)\left[\mathrm{1}+\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} }\left(\Phi+\frac{\mathrm{1}}{\mathrm{4}}\right)\right]^{\mathrm{2}} =\frac{\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{s}^{\mathrm{2}} } \\ $$$${a}^{\mathrm{2}} \left(\mathrm{2}\Phi+\mathrm{1}\right)\left[\mathrm{4}{b}^{\mathrm{2}} +\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left(\mathrm{4}\Phi−\mathrm{1}\right)\right]^{\mathrm{2}} +{b}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}\Phi\right)\left[\mathrm{4}{a}^{\mathrm{2}} +\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left(\mathrm{4}\Phi+\mathrm{1}\right)\right]^{\mathrm{2}} =\frac{\mathrm{96}{a}^{\mathrm{4}} {b}^{\mathrm{4}} }{{s}^{\mathrm{2}} } \\ $$$$\mathrm{32}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{3}} \Phi^{\mathrm{3}} −\mathrm{6}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{3}} \Phi+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left[\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{12}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right]−\frac{\mathrm{96}{a}^{\mathrm{4}} {b}^{\mathrm{4}} }{{s}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left[\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\Phi\right]^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{16}}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} \left[\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\Phi\right]+\left\{\frac{\mathrm{1}}{\mathrm{32}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left[\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{12}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right]−\frac{\mathrm{3}{a}^{\mathrm{4}} {b}^{\mathrm{4}} }{{s}^{\mathrm{2}} }\right\}=\mathrm{0} \\ $$$${with}\:\mu=\frac{{b}}{{a}}<\mathrm{1},\:\xi=\frac{{s}}{{a}} \\ $$$$\left[\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\Phi\right]^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{16}}\left(\mathrm{1}−\mu^{\mathrm{2}} \right)^{\mathrm{2}} \left[\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\Phi\right]+\left\{\frac{\mathrm{1}}{\mathrm{32}}\left(\mathrm{1}+\mu^{\mathrm{2}} \right)\left[\left(\mathrm{1}+\mu^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{12}\mu^{\mathrm{2}} \right]−\frac{\mathrm{3}\mu^{\mathrm{4}} }{\xi^{\mathrm{2}} }\right\}=\mathrm{0} \\ $$$${with} \\ $$$${z}=\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\Phi=\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\left(\mathrm{cos}^{\mathrm{2}} \:\theta−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${u}=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}−\mu^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${v}=\frac{\mathrm{1}}{\mathrm{64}}\left(\mathrm{1}+\mu^{\mathrm{2}} \right)\left[\left(\mathrm{1}+\mu^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{12}\mu^{\mathrm{2}} \right]−\frac{\mathrm{3}\mu^{\mathrm{4}} }{\mathrm{2}\xi^{\mathrm{2}} } \\ $$$$\Rightarrow{z}^{\mathrm{3}} −\mathrm{3}{uz}+\mathrm{2}{v}=\mathrm{0} \\ $$$$\Delta=−{u}^{\mathrm{3}} +{v}^{\mathrm{2}} <\mathrm{0} \\ $$$${z}=\mathrm{2}\sqrt{{u}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{v}}{\sqrt{{u}^{\mathrm{3}} }}+\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right) \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}\sqrt{\mu}}{\mathrm{1}−\mu^{\mathrm{2}} }\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{v}}{\sqrt{{u}^{\mathrm{3}} }}+\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right)} \\ $$ | ||
Commented by ajfour last updated on 03/Jul/19 | ||
$${Now}\:{its}\:{beauty},\:{Sir}! \\ $$$${Does}\:{it}\:{imply},\:{only}\:{one}\:{real}\:{root}\:{for} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \theta\:\:,\:{a},{b},{s}\:{being}\:{given}. \\ $$ | ||
Commented by ajfour last updated on 02/Jul/19 | ||
$${your}\:{faith}\:{is}\:{beyond}\:{all}\:{realms}! \\ $$ | ||
Commented by ajfour last updated on 02/Jul/19 | ||
$${Please}\:{explain}\:{what}\:{is}\:{D}\:{and} \\ $$$${how}\:{you}\:{express}\:{it},\:{Sir}.. \\ $$ | ||
Commented by mr W last updated on 02/Jul/19 | ||
$${thank}\:{you}\:{for}\:{following}\:{sir}! \\ $$$${with}\:{D}\:{i}\:{just}\:{don}'{t}\:{want}\:{to}\:{write} \\ $$$${too}\:{much}\:{when}\:{solving}\:{the}\:{equation} \\ $$$${system}. \\ $$ | ||
Commented by mr W last updated on 02/Jul/19 | ||
Commented by mr W last updated on 03/Jul/19 | ||
$$\frac{\mathrm{3}\mu^{\mathrm{4}} }{\xi^{\mathrm{2}} }=\left(\mathrm{1}−\mu^{\mathrm{2}} \right)^{\mathrm{3}} \Phi^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{16}}\left(\mathrm{1}−\mu^{\mathrm{2}} \right)^{\mathrm{3}} \Phi+\frac{\mathrm{1}}{\mathrm{32}}\left(\mathrm{1}+\mu^{\mathrm{2}} \right)^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{8}}\mu^{\mathrm{2}} \left(\mathrm{1}+\mu^{\mathrm{2}} \right) \\ $$$$\frac{{d}}{{d}\Phi}\left(\frac{\mathrm{3}\mu^{\mathrm{4}} }{\xi^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}\Phi^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{16}}=\mathrm{0} \\ $$$$\Rightarrow\Phi=\pm\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\mathrm{cos}^{\mathrm{2}} \:\theta=\pm\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{cos}\:\theta=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}},\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{max}.\:{s}\:{or}\:{min}.\:{s}\:{at}\:\theta=\frac{{n}\pi}{\mathrm{2}}\pm\frac{\pi}{\mathrm{6}} \\ $$$${with}\:\Phi=\frac{\mathrm{1}}{\mathrm{4}}: \\ $$$$\frac{\mathrm{3}\mu^{\mathrm{4}} }{\xi^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{32}}\left(\mathrm{1}−\mu^{\mathrm{2}} \right)^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{32}}\left(\mathrm{1}+\mu^{\mathrm{2}} \right)^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{8}}\mu^{\mathrm{2}} \left(\mathrm{1}+\mu^{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{3}\mu^{\mathrm{2}} }{\xi^{\mathrm{2}} }=\frac{\left(\mathrm{3}+\mu^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\Rightarrow\xi^{\mathrm{2}} =\frac{\mathrm{48}\mu^{\mathrm{2}} }{\left(\mathrm{3}+\mu^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\xi_{{max}} =\frac{\mathrm{4}\sqrt{\mathrm{3}}\mu}{\mathrm{3}+\mu^{\mathrm{2}} }=\frac{{s}_{{max}} }{{a}} \\ $$$${with}\:\Phi=−\frac{\mathrm{1}}{\mathrm{4}}: \\ $$$$\frac{\mathrm{3}\mu^{\mathrm{4}} }{\xi^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{32}}\left(\mathrm{1}−\mu^{\mathrm{2}} \right)^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{32}}\left(\mathrm{1}+\mu^{\mathrm{2}} \right)^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{8}}\mu^{\mathrm{2}} \left(\mathrm{1}+\mu^{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{3}\mu^{\mathrm{4}} }{\xi^{\mathrm{2}} }=\frac{\left(\mathrm{1}+\mathrm{3}\mu^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\Rightarrow\xi^{\mathrm{2}} =\frac{\mathrm{48}\mu^{\mathrm{4}} }{\left(\mathrm{1}+\mathrm{3}\mu^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\xi_{{min}} =\frac{\mathrm{4}\sqrt{\mathrm{3}}\mu^{\mathrm{2}} }{\mathrm{1}+\mathrm{3}\mu^{\mathrm{2}} }=\frac{{s}_{{min}} }{{a}} \\ $$ | ||
Answered by ajfour last updated on 06/Jul/19 | ||
$${let}\:{mid}-{point}\:{of}\:{BC}\:{be}\:{M}\left({h},{k}\right). \\ $$$${slope}\:{of}\:{AM}\:{be}\:{m}=\mathrm{tan}\:\theta \\ $$$${x}_{{A}} ={h}+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\theta,\:\:\:{y}_{{A}} ={k}+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\theta \\ $$$${x}_{{B}} ={h}−\frac{{s}}{\mathrm{2}}\mathrm{sin}\:\theta\:\:,\:\:{y}_{{B}} ={k}+\frac{{s}}{\mathrm{2}}\mathrm{cos}\:\theta \\ $$$${x}_{{C}} ={h}+\frac{{s}}{\mathrm{2}}\mathrm{sin}\:\theta\:\:,\:\:{y}_{{C}} ={k}−\frac{{s}}{\mathrm{2}}\mathrm{cos}\:\theta \\ $$$${As}\:{B},{C}\:\:{lie}\:{on}\:{ellipse} \\ $$$$\:\:\:\:\frac{{x}_{{B}} ^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}_{{B}} ^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:,\:\:\frac{{x}_{{C}} ^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}_{{C}} ^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${subtracting} \\ $$$$\frac{\left({x}_{{B}} −{x}_{{C}} \right)\left({x}_{{B}} +{x}_{{C}} \right)}{{a}^{\mathrm{2}} }+\frac{\left({y}_{{B}} −{y}_{{C}} \right)\left({y}_{{B}} +{y}_{{C}} \right)}{{b}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\:−\mathrm{2}{b}^{\mathrm{2}} {sh}\mathrm{sin}\:\theta+\mathrm{2}{a}^{\mathrm{2}} {sk}\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{k}=\frac{{b}^{\mathrm{2}} {h}}{{a}^{\mathrm{2}} }\mathrm{tan}\:\theta \\ $$$${similarly}\:{as}\:{A},{B}\:{lie}\:{on}\:{ellipse} \\ $$$$\Rightarrow\:\frac{{b}^{\mathrm{2}} {s}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\left[\mathrm{2}{h}+\frac{{s}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)\right] \\ $$$$\:+\frac{{a}^{\mathrm{2}} {s}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−\mathrm{cos}\:\theta\right)\left[\mathrm{2}{k}+\frac{{s}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)\right]=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}{hb}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right) \\ $$$$+\mathrm{2}{b}^{\mathrm{2}} {h}\left(\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right)\left(\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−\mathrm{cos}\:\theta\right) \\ $$$$+\frac{{s}}{\mathrm{2}}\left\{{b}^{\mathrm{2}} \left(\mathrm{3cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\right)+{a}^{\mathrm{2}} \left(\mathrm{3sin}\:^{\mathrm{2}} \theta−\mathrm{cos}\:^{\mathrm{2}} \theta\right)\right\}=\mathrm{0} \\ $$$$\Rightarrow\:{h}=−\frac{{s}\mathrm{cos}\:\theta}{\mathrm{4}\sqrt{\mathrm{3}}{b}^{\mathrm{2}} }\left\{{a}^{\mathrm{2}} \left(\mathrm{3sin}\:^{\mathrm{2}} \theta−\mathrm{cos}\:^{\mathrm{2}} \theta\right)+{b}^{\mathrm{2}} \left(\mathrm{3cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\right)\right\} \\ $$$${And}\:{as}\:{A}\:{lies}\:{on}\:{ellipse}, \\ $$$$\:\:\:{b}^{\mathrm{2}} \left({h}+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\frac{{b}^{\mathrm{2}} {h}}{{a}^{\mathrm{2}} }\mathrm{tan}\:\theta+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\theta\right)^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\Rightarrow\:{b}^{\mathrm{2}} {h}^{\mathrm{2}} +{b}^{\mathrm{2}} {sh}\sqrt{\mathrm{3}}\mathrm{cos}\:\theta+\frac{\mathrm{3}{b}^{\mathrm{2}} {s}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{4}} \\ $$$$\:\:\:+\frac{{b}^{\mathrm{4}} {h}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta}{{a}^{\mathrm{2}} }+{b}^{\mathrm{2}} {sh}\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\mathrm{tan}\:\theta \\ $$$$\:\:\:+\frac{\mathrm{3}{a}^{\mathrm{2}} {s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{4}}={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\Rightarrow\:{b}^{\mathrm{2}} {h}^{\mathrm{2}} \left(\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\mathrm{tan}\:^{\mathrm{2}} \theta\right)+\frac{{b}^{\mathrm{2}} {sh}\sqrt{\mathrm{3}}}{\mathrm{cos}\:\theta} \\ $$$$\:\:\:\:+\frac{\mathrm{3}{s}^{\mathrm{2}} }{\mathrm{4}}\left({b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta+{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta\right)={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\left({a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta+{b}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta\right)\frac{{h}^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} \theta} \\ $$$$\:\:+\left({b}^{\mathrm{2}} {s}\sqrt{\mathrm{3}}\right)\frac{{h}}{\mathrm{cos}\:\theta}+\frac{\mathrm{3}}{\mathrm{4}}{s}^{\mathrm{2}} \left({b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta+{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta\right) \\ $$$$\:\:\:\:\:\:\:\:−{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:\:\mathrm{cos}\:\theta={t}\:,\:{then} \\ $$$$\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\left(\frac{{h}}{{t}}\right)^{\mathrm{2}} \left\{{b}^{\mathrm{2}} +\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){t}^{\mathrm{2}} \right\}+{b}^{\mathrm{2}} {s}\sqrt{\mathrm{3}}\left(\frac{{h}}{{t}}\right) \\ $$$$\:\:\:+\frac{\mathrm{3}{s}^{\mathrm{2}} }{\mathrm{4}}\left[{a}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){t}^{\mathrm{2}} \right]\:−{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$${where} \\ $$$$\:{h}=−\frac{{s}\mathrm{cos}\:\theta}{\mathrm{4}\sqrt{\mathrm{3}}{b}^{\mathrm{2}} }\left\{{a}^{\mathrm{2}} \left(\mathrm{3sin}\:^{\mathrm{2}} \theta−\mathrm{cos}\:^{\mathrm{2}} \theta\right)+{b}^{\mathrm{2}} \left(\mathrm{3cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\right)\right\} \\ $$$$\Rightarrow\:\frac{{h}}{{t}}=−\frac{{s}}{\mathrm{4}\sqrt{\mathrm{3}}{b}^{\mathrm{2}} }\left\{\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{4}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){t}^{\mathrm{2}} \right\} \\ $$$${Now} \\ $$$$\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\left(\frac{{s}^{\mathrm{2}} }{\mathrm{48}{b}^{\mathrm{4}} }\right)\left\{{b}^{\mathrm{2}} +\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){t}^{\mathrm{2}} \right\}\left\{\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{4}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){t}^{\mathrm{2}} \right\}^{\mathrm{2}} \\ $$$$\:−{b}^{\mathrm{2}} {s}\sqrt{\mathrm{3}}\left(\frac{{s}}{\mathrm{4}\sqrt{\mathrm{3}}{b}^{\mathrm{2}} }\right)\left\{\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{4}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){t}^{\mathrm{2}} \right\} \\ $$$$\:\:\:+\frac{\mathrm{3}{s}^{\mathrm{2}} }{\mathrm{4}}\left[{a}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){t}^{\mathrm{2}} \right]\:−{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\:{let}\:\:\:\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){t}^{\mathrm{2}} ={v} \\ $$$$\frac{{s}^{\mathrm{2}} }{\mathrm{48}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\left({b}^{\mathrm{2}} +{v}\right)\left(\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{4}{v}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\frac{{s}^{\mathrm{2}} }{\mathrm{4}}\left({b}^{\mathrm{2}} +{v}\right)−{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:\:{b}^{\mathrm{2}} +{v}=\:{y} \\ $$$$\Rightarrow\:\left(\frac{{s}^{\mathrm{2}} }{\mathrm{48}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\right){y}\left\{\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{4}{y}\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{{s}^{\mathrm{2}} {y}}{\mathrm{4}}−{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:{s}^{\mathrm{2}} {y}\left\{\left[\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{4}{y}\right]^{\mathrm{2}} +\mathrm{12}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{48}{a}^{\mathrm{4}} {b}^{\mathrm{4}} =\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$ \\ $$ | ||
Commented by mr W last updated on 01/Jul/19 | ||
$${great}\:{solution}\:{sir}! \\ $$ | ||
Commented by mr W last updated on 03/Jul/19 | ||
$${sir},\:{please}\:{check}: \\ $$$${s}_{\mathrm{0}} ^{\mathrm{2}} \:=\frac{\mathrm{48}{r}^{\mathrm{4}} {a}^{\mathrm{2}} }{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)\left[\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{3}\right]}\:\Rightarrow{should}\:{be}\:{min}.{s} \\ $$$${s}_{{min}} =\frac{\mathrm{4}\sqrt{\mathrm{3}}{b}}{\mathrm{3}+\left(\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)}\:\Rightarrow\:{should}\:{be}\:{max}.\:{s}. \\ $$ | ||
Commented by mr W last updated on 03/Jul/19 | ||
$${an}\:{other}\:{question}\:{is}: \\ $$$${why}\:\frac{{b}}{{a}}\in\left(\mathrm{0},\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right]\:?\:{not}\:\frac{{b}}{{a}}\in\left(\mathrm{0},\mathrm{1}\right)\:? \\ $$$$\left({assume}\:{b}<{a}\right) \\ $$ | ||
Commented by mr W last updated on 06/Jul/19 | ||
$${i}\:{found}\:{out}:\: \\ $$$${max}.\:{s}\:{is}\:{always}\:{at}\:\theta=\mathrm{30}°\:{or}\:\mathrm{90}° \\ $$$${min}.\:{s}\:{is}\:{always}\:{at}\:\theta=\mathrm{0}°\:{or}\:\mathrm{60}° \\ $$ | ||
Commented by mr W last updated on 03/Jul/19 | ||
$${with}\:{r}\leqslant\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:{something}\:{must}\:{be}\:{wrong}. \\ $$$${in}\:{fact}\:{there}\:{is}\:{solution}\:{for}\:{any} \\ $$$$\mathrm{0}<{r}<\mathrm{1}. \\ $$$${eg}.\:{a}=\mathrm{4},\:{b}=\mathrm{3},\:\Rightarrow{r}=\mathrm{0}.\mathrm{75}>\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}: \\ $$$${at}\:\theta=\mathrm{0}°:\:{s}=\mathrm{5}.\mathrm{8344} \\ $$$${at}\:\theta=\mathrm{30}°:\:{s}=\mathrm{5}.\mathrm{8004} \\ $$$${at}\:\theta=\mathrm{45}°:\:{s}=\mathrm{5}.\mathrm{8172} \\ $$$${with}\:\theta\:{as}\:{meant}\:{in}\:{your}\:{solution}. \\ $$ | ||
Commented by mr W last updated on 06/Jul/19 | ||
Commented by mr W last updated on 06/Jul/19 | ||
$${this}\:{is}\:{the}\:\theta\:{in}\:{my}\:{solution}. \\ $$ | ||