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Question-46611




Question Number 46611 by ajfour last updated on 29/Oct/18
Commented by ajfour last updated on 29/Oct/18
A ring of radius R stands upright  at G(a,b,0). Along which line, in  xy-plane and passing through G,  as axis should the ring be tilted  and by how much (θ) such that  it comes in contact with xz, and  yz plane simultaneously ?
$${A}\:{ring}\:{of}\:{radius}\:{R}\:{stands}\:{upright} \\ $$$${at}\:{G}\left({a},{b},\mathrm{0}\right).\:{Along}\:{which}\:{line},\:{in} \\ $$$${xy}-{plane}\:{and}\:{passing}\:{through}\:{G}, \\ $$$${as}\:{axis}\:{should}\:{the}\:{ring}\:{be}\:{tilted} \\ $$$${and}\:{by}\:{how}\:{much}\:\left(\theta\right)\:{such}\:{that} \\ $$$${it}\:{comes}\:{in}\:{contact}\:{with}\:{xz},\:{and} \\ $$$${yz}\:{plane}\:{simultaneously}\:? \\ $$
Answered by MrW3 last updated on 29/Oct/18
Commented by MrW3 last updated on 30/Oct/18
ΔABC represents the plane which the  circle lies in. The circle is the inscribed  circle of the triangle ΔABC.  A(a,0,0)  B(0,b,0)  C(0,0,c)  we need to determine a,b and c.    G(h,k,0)  let ∠OAB=α  ⇒AG=(k/(sin α)), GB=(h/(cos α))  AB=AG+GB  ⇒AB=(k/(sin α))+(h/(cos α))  tan (A/2)=(R/(AG))=((R sin α)/k)  tan (B/2)=(R/(GB))=((R cos α)/h)  AC=AP+PC=AG+(R/(tan (C/2)))=(k/(sin α))+(R/(tan ((π−A−B)/2)))  AC=(k/(sin α))+R tan ((A/2)+(B/2))=(k/(sin α))+R((((R sin α)/k)+((R cos α)/h))/(1−((R^2  sin α cos α)/(hk))))  ⇒AC=(k/(sin α))+((R^2 (h sin α+k cos α))/(hk−R^2  sin α cos α))  BC=BQ+QC=GB+PC  ⇒BC=(h/(cos α))+((R^2 (h sin α+k cos α))/(hk−R^2  sin α cos α))    AB^2 =a^2 +b^2   AC^2 =a^2 +c^2   BC^2 =b^2 +c^2   ⇒a^2 +b^2 +c^2 =((AB^2 +AC^2 +BC^2 )/2)  ⇒a^2 =((AB^2 +AC^2 −BC^2 )/2)=(([(k/(sin α))+(h/(cos α))]^2 +[(k/(sin α))+((R^2 (h sin α+k cos α))/(hk−R^2  sin α cos α))]^2 −[(h/(cos α))+((R^2 (h sin α+k cos α))/(hk−R^2  sin α cos α))]^2 )/2)  ...(i)  ⇒b^2 =((AB^2 −AC^2 +BC^2 )/2)=(([(k/(sin α))+(h/(cos α))]^2 −[(k/(sin α))+((R^2 (h sin α+k cos α))/(hk−R^2  sin α cos α))]^2 +[(h/(cos α))+((R^2 (h sin α+k cos α))/(hk−R^2  sin α cos α))]^2 )/2)    ...(ii)  ⇒c^2 =((−AB^2 +AC^2 +BC^2 )/2)=((−[(k/(sin α))+(h/(cos α))]^2 +[(k/(sin α))+((R^2 (h sin α+k cos α))/(hk−R^2  sin α cos α))]^2 +[(h/(cos α))+((R^2 (h sin α+k cos α))/(hk−R^2  sin α cos α))]^2 )/2)   ...(iii)    on the other side we have  AC^2 −(AB cos α)^2 =BC^2 −(AB sin α)^2   AC^2 −BC^2 =AB^2  (cos^2  α−sin^2  α)=cos 2α AB^2   ⇒[(k/(sin α))+((R^2 (h sin α+k cos α))/(hk−R^2  sin α cos α))]^2 −[(h/(cos α))+((R^2 (h sin α+k cos α))/(hk−R^2  sin α cos α))]^2 =cos 2α [(k/(sin α))+(h/(cos α))]^2   ⇒[(k/(sin α))+(h/(cos α))+((2R^2 (h sin α+k cos α))/(hk−R^2  sin α cos α))][(k/(sin α))−(h/(cos α))]=cos 2α [(k/(sin α))+(h/(cos α))]^2   ⇒(((2hk)/R^2 )+sin 2α)((k/h)−tan α)=(((2hk)/R^2 )−sin 2α)((k/h)+tan α)cos 2α   ...(iv)  let t=tan α, μ=(k/h), λ=(R^2 /(hk))  ⇒(μ−λ)t^3 −t^2 +μt+(μλ−1)=0   ...(v)    from (v) we get α, with this α we can  get a,b and c from (i) to (iii).    let d=altitude from O to AB  (1/2)d×AB=(1/2)ab⇒d=((ab)/(AB))=((ab)/((k/(sin α))+(h/(cos α))))=((ab sin α cos α)/(k cos α+h sin α))  tan θ=(d/c)  ⇒θ=tan^(−1) ((ab sin α cos α)/(c(k cos α+h sin α)))
$$\Delta{ABC}\:{represents}\:{the}\:{plane}\:{which}\:{the} \\ $$$${circle}\:{lies}\:{in}.\:{The}\:{circle}\:{is}\:{the}\:{inscribed} \\ $$$${circle}\:{of}\:{the}\:{triangle}\:\Delta{ABC}. \\ $$$${A}\left({a},\mathrm{0},\mathrm{0}\right) \\ $$$${B}\left(\mathrm{0},{b},\mathrm{0}\right) \\ $$$${C}\left(\mathrm{0},\mathrm{0},{c}\right) \\ $$$${we}\:{need}\:{to}\:{determine}\:{a},{b}\:{and}\:{c}. \\ $$$$ \\ $$$${G}\left({h},{k},\mathrm{0}\right) \\ $$$${let}\:\angle{OAB}=\alpha \\ $$$$\Rightarrow{AG}=\frac{{k}}{\mathrm{sin}\:\alpha},\:{GB}=\frac{{h}}{\mathrm{cos}\:\alpha} \\ $$$${AB}={AG}+{GB} \\ $$$$\Rightarrow{AB}=\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{h}}{\mathrm{cos}\:\alpha} \\ $$$$\mathrm{tan}\:\frac{{A}}{\mathrm{2}}=\frac{{R}}{{AG}}=\frac{{R}\:\mathrm{sin}\:\alpha}{{k}} \\ $$$$\mathrm{tan}\:\frac{{B}}{\mathrm{2}}=\frac{{R}}{{GB}}=\frac{{R}\:\mathrm{cos}\:\alpha}{{h}} \\ $$$${AC}={AP}+{PC}={AG}+\frac{{R}}{\mathrm{tan}\:\frac{{C}}{\mathrm{2}}}=\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{R}}{\mathrm{tan}\:\frac{\pi−{A}−{B}}{\mathrm{2}}} \\ $$$${AC}=\frac{{k}}{\mathrm{sin}\:\alpha}+{R}\:\mathrm{tan}\:\left(\frac{{A}}{\mathrm{2}}+\frac{{B}}{\mathrm{2}}\right)=\frac{{k}}{\mathrm{sin}\:\alpha}+{R}\frac{\frac{{R}\:\mathrm{sin}\:\alpha}{{k}}+\frac{{R}\:\mathrm{cos}\:\alpha}{{h}}}{\mathrm{1}−\frac{{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{{hk}}} \\ $$$$\Rightarrow{AC}=\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{R}^{\mathrm{2}} \left({h}\:\mathrm{sin}\:\alpha+{k}\:\mathrm{cos}\:\alpha\right)}{{hk}−{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha} \\ $$$${BC}={BQ}+{QC}={GB}+{PC} \\ $$$$\Rightarrow{BC}=\frac{{h}}{\mathrm{cos}\:\alpha}+\frac{{R}^{\mathrm{2}} \left({h}\:\mathrm{sin}\:\alpha+{k}\:\mathrm{cos}\:\alpha\right)}{{hk}−{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha} \\ $$$$ \\ $$$${AB}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${AC}^{\mathrm{2}} ={a}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$${BC}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\frac{{AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} +{BC}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{{AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} −{BC}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left[\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{h}}{\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} +\left[\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{R}^{\mathrm{2}} \left({h}\:\mathrm{sin}\:\alpha+{k}\:\mathrm{cos}\:\alpha\right)}{{hk}−{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} −\left[\frac{{h}}{\mathrm{cos}\:\alpha}+\frac{{R}^{\mathrm{2}} \left({h}\:\mathrm{sin}\:\alpha+{k}\:\mathrm{cos}\:\alpha\right)}{{hk}−{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} }{\mathrm{2}}\:\:…\left({i}\right) \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\frac{{AB}^{\mathrm{2}} −{AC}^{\mathrm{2}} +{BC}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left[\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{h}}{\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} −\left[\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{R}^{\mathrm{2}} \left({h}\:\mathrm{sin}\:\alpha+{k}\:\mathrm{cos}\:\alpha\right)}{{hk}−{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} +\left[\frac{{h}}{\mathrm{cos}\:\alpha}+\frac{{R}^{\mathrm{2}} \left({h}\:\mathrm{sin}\:\alpha+{k}\:\mathrm{cos}\:\alpha\right)}{{hk}−{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} }{\mathrm{2}}\:\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow{c}^{\mathrm{2}} =\frac{−{AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} +{BC}^{\mathrm{2}} }{\mathrm{2}}=\frac{−\left[\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{h}}{\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} +\left[\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{R}^{\mathrm{2}} \left({h}\:\mathrm{sin}\:\alpha+{k}\:\mathrm{cos}\:\alpha\right)}{{hk}−{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} +\left[\frac{{h}}{\mathrm{cos}\:\alpha}+\frac{{R}^{\mathrm{2}} \left({h}\:\mathrm{sin}\:\alpha+{k}\:\mathrm{cos}\:\alpha\right)}{{hk}−{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} }{\mathrm{2}}\:\:\:…\left({iii}\right) \\ $$$$ \\ $$$${on}\:{the}\:{other}\:{side}\:{we}\:{have} \\ $$$${AC}^{\mathrm{2}} −\left({AB}\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} ={BC}^{\mathrm{2}} −\left({AB}\:\mathrm{sin}\:\alpha\right)^{\mathrm{2}} \\ $$$${AC}^{\mathrm{2}} −{BC}^{\mathrm{2}} ={AB}^{\mathrm{2}} \:\left(\mathrm{cos}^{\mathrm{2}} \:\alpha−\mathrm{sin}^{\mathrm{2}} \:\alpha\right)=\mathrm{cos}\:\mathrm{2}\alpha\:{AB}^{\mathrm{2}} \\ $$$$\Rightarrow\left[\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{R}^{\mathrm{2}} \left({h}\:\mathrm{sin}\:\alpha+{k}\:\mathrm{cos}\:\alpha\right)}{{hk}−{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} −\left[\frac{{h}}{\mathrm{cos}\:\alpha}+\frac{{R}^{\mathrm{2}} \left({h}\:\mathrm{sin}\:\alpha+{k}\:\mathrm{cos}\:\alpha\right)}{{hk}−{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} =\mathrm{cos}\:\mathrm{2}\alpha\:\left[\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{h}}{\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} \\ $$$$\Rightarrow\left[\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{h}}{\mathrm{cos}\:\alpha}+\frac{\mathrm{2}{R}^{\mathrm{2}} \left({h}\:\mathrm{sin}\:\alpha+{k}\:\mathrm{cos}\:\alpha\right)}{{hk}−{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\right]\left[\frac{{k}}{\mathrm{sin}\:\alpha}−\frac{{h}}{\mathrm{cos}\:\alpha}\right]=\mathrm{cos}\:\mathrm{2}\alpha\:\left[\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{h}}{\mathrm{cos}\:\alpha}\right]^{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{\mathrm{2}{hk}}{{R}^{\mathrm{2}} }+\mathrm{sin}\:\mathrm{2}\alpha\right)\left(\frac{{k}}{{h}}−\mathrm{tan}\:\alpha\right)=\left(\frac{\mathrm{2}{hk}}{{R}^{\mathrm{2}} }−\mathrm{sin}\:\mathrm{2}\alpha\right)\left(\frac{{k}}{{h}}+\mathrm{tan}\:\alpha\right)\mathrm{cos}\:\mathrm{2}\alpha\:\:\:…\left({iv}\right) \\ $$$${let}\:{t}=\mathrm{tan}\:\alpha,\:\mu=\frac{{k}}{{h}},\:\lambda=\frac{{R}^{\mathrm{2}} }{{hk}} \\ $$$$\Rightarrow\left(\mu−\lambda\right){t}^{\mathrm{3}} −{t}^{\mathrm{2}} +\mu{t}+\left(\mu\lambda−\mathrm{1}\right)=\mathrm{0}\:\:\:…\left({v}\right) \\ $$$$ \\ $$$${from}\:\left({v}\right)\:{we}\:{get}\:\alpha,\:{with}\:{this}\:\alpha\:{we}\:{can} \\ $$$${get}\:{a},{b}\:{and}\:{c}\:{from}\:\left({i}\right)\:{to}\:\left({iii}\right). \\ $$$$ \\ $$$${let}\:{d}={altitude}\:{from}\:{O}\:{to}\:{AB} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{d}×{AB}=\frac{\mathrm{1}}{\mathrm{2}}{ab}\Rightarrow{d}=\frac{{ab}}{{AB}}=\frac{{ab}}{\frac{{k}}{\mathrm{sin}\:\alpha}+\frac{{h}}{\mathrm{cos}\:\alpha}}=\frac{{ab}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{{k}\:\mathrm{cos}\:\alpha+{h}\:\mathrm{sin}\:\alpha} \\ $$$$\mathrm{tan}\:\theta=\frac{{d}}{{c}} \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \frac{{ab}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{{c}\left({k}\:\mathrm{cos}\:\alpha+{h}\:\mathrm{sin}\:\alpha\right)} \\ $$
Commented by MrW3 last updated on 30/Oct/18
Commented by ajfour last updated on 30/Oct/18
GREAT  solution Sir,  Thanks !
$$\mathcal{GREAT}\:\:{solution}\:{Sir}, \\ $$$$\mathcal{T}{hanks}\:! \\ $$
Commented by MrW3 last updated on 30/Oct/18
Commented by MrW3 last updated on 30/Oct/18
ajfour sir: can you confirm the result  via other method? I have a question  for example: with a given point G(h,k)  what is the range for R such that the  action is possible?
$${ajfour}\:{sir}:\:{can}\:{you}\:{confirm}\:{the}\:{result} \\ $$$${via}\:{other}\:{method}?\:{I}\:{have}\:{a}\:{question} \\ $$$${for}\:{example}:\:{with}\:{a}\:{given}\:{point}\:{G}\left({h},{k}\right) \\ $$$${what}\:{is}\:{the}\:{range}\:{for}\:{R}\:{such}\:{that}\:{the} \\ $$$${action}\:{is}\:{possible}? \\ $$
Commented by MrW3 last updated on 30/Oct/18
Commented by MrW3 last updated on 30/Oct/18
R_0 =(√(h^2 +k^2 ))  if R=R_0  the ring can only stand upright  and can not be tilted towards the walls.  (h−R_(min) )^2 +(k−R_(min) )^2 =R_(min) ^2   R_(min) ^2 −2(h+k)R_(min) +(h^2 +k^2 )=0  ⇒R_(min) =h+k±(√(2hk))  ⇒R_(min) =h+k−(√(2hk))  ⇒R_(max) =h+k+(√(2hk))    case 1: R_(min) ≤R≤R_0   the ring can be tilted towards  the walls and touch both.  case 2: R_0 <R≤R_(max)   the ring can touch the walls only from  outside.
$${R}_{\mathrm{0}} =\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} } \\ $$$${if}\:{R}={R}_{\mathrm{0}} \:{the}\:{ring}\:{can}\:{only}\:{stand}\:{upright} \\ $$$${and}\:{can}\:{not}\:{be}\:{tilted}\:{towards}\:{the}\:{walls}. \\ $$$$\left({h}−{R}_{{min}} \right)^{\mathrm{2}} +\left({k}−{R}_{{min}} \right)^{\mathrm{2}} ={R}_{{min}} ^{\mathrm{2}} \\ $$$${R}_{{min}} ^{\mathrm{2}} −\mathrm{2}\left({h}+{k}\right){R}_{{min}} +\left({h}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{R}_{{min}} ={h}+{k}\pm\sqrt{\mathrm{2}{hk}} \\ $$$$\Rightarrow{R}_{{min}} ={h}+{k}−\sqrt{\mathrm{2}{hk}} \\ $$$$\Rightarrow{R}_{{max}} ={h}+{k}+\sqrt{\mathrm{2}{hk}} \\ $$$$ \\ $$$${case}\:\mathrm{1}:\:{R}_{{min}} \leqslant{R}\leqslant{R}_{\mathrm{0}} \\ $$$${the}\:{ring}\:{can}\:{be}\:{tilted}\:{towards} \\ $$$${the}\:{walls}\:{and}\:{touch}\:{both}. \\ $$$${case}\:\mathrm{2}:\:{R}_{\mathrm{0}} <{R}\leqslant{R}_{{max}} \\ $$$${the}\:{ring}\:{can}\:{touch}\:{the}\:{walls}\:{only}\:{from} \\ $$$${outside}. \\ $$
Commented by ajfour last updated on 31/Oct/18
I couldn′t find any good way , Sir  and was busy unentangling my  hands!
$${I}\:{couldn}'{t}\:{find}\:{any}\:{good}\:{way}\:,\:{Sir} \\ $$$${and}\:{was}\:{busy}\:{unentangling}\:{my} \\ $$$${hands}! \\ $$
Commented by MrW3 last updated on 31/Oct/18
thank you for responding sir! this is  a very nice and challenging  question.
$${thank}\:{you}\:{for}\:{responding}\:{sir}!\:{this}\:{is} \\ $$$${a}\:{very}\:{nice}\:{and}\:{challenging}\:\:{question}. \\ $$
Commented by ajfour last updated on 31/Oct/18
i think this is same as previous   one, Sir.  By the way you were  too pragmatic in suggesting  to  obtain tan θ= (d/c) , Sir.
$${i}\:{think}\:{this}\:{is}\:{same}\:{as}\:{previous}\: \\ $$$${one},\:{Sir}.\:\:{By}\:{the}\:{way}\:{you}\:{were} \\ $$$${too}\:{pragmatic}\:{in}\:{suggesting}\:\:{to} \\ $$$${obtain}\:\mathrm{tan}\:\theta=\:\frac{{d}}{{c}}\:,\:{Sir}. \\ $$
Answered by MrW3 last updated on 01/Nov/18
Commented by MrW3 last updated on 01/Nov/18
another (better) way to solve:    the projection of the circle in the  xy−plane is an ellipse which has the  semi−major axis R and tangents  x−axis and y−axis simultaneously.    assume the semi−minor axis of ellipse  is d.  in coordinate system x′y′ the eqn.  of ellipse is  (x^2 /R^2 )+(y^2 /d^2 )=1  GA=(k/(sin α)), GB=(h/(cos α))    eqn. of line OA in x′y′ system is  ((y−d)/(x−(k/(sin α))))=tan α  ⇒sin α x−cos α y+d cos α−k=0  since it′s a tangent line of ellipse,  R^2  sin^2  α+d^2  cos^2  α−(d cos α−k)^2 =0  ⇒R^2  sin^2  α−k^2 +2dk cos α=0   ...(i)    eqn. of line OB in x′y′ system is  ((y−d)/(x+(h/(cos α))))=−(1/(tan α))  ⇒cos α x+sin α y+h−d sin α=0  since it′s a tangent line of ellipse,  R^2  cos^2  α+q^2  sin^2  α−(h−d sin α)^2 =0  ⇒R^2  cos^2  α−h^2 +2dh sin α=0   ...(ii)    (i)+(ii):  R^2 −k^2 −h^2 +2d(k cos α+h sin α)=0  ⇒d=((k^2 +h^2 −R^2 )/(2(k cos α+h sin α)))   ...(iii)  put this into (ii):  R^2  cos^2  α+((h(k^2 +h^2 −R^2 )sin α)/(k cos α+h sin α))=h^2   ⇒((R/h))^2 (1/(1+tan^2  α))+((1+((k/h))^2 −((R/h))^2 )/(1+((k/h))(1/(tan α))))=1  ...(iv)  or with t=tan α, λ=(R/h), μ=(k/h)  ⇒(μ^2 −λ^2 )t^3 −μt^2 +μ^2 t+μ(λ^2 −1)=0   ...(iv)    from (iv) we can get α at first.  then we get d from (iii).    sin θ=(d/R)=((k^2 +h^2 −R^2 )/(2R(k cos α+h sin α)))  ⇒θ=sin^(−1) ((1+((k/h))^2 −((R/h))^2 )/(2((R/h))[sin α+((k/h))cos  α]))    examples:  h=4, k=3, R=4:  ⇒α=0°, 29.4346°  ⇒θ=22.024°, 14.225°    h=3, k=3, R=2.5:  ⇒α=30.8212°, 45°, 59.1788°  ⇒θ=34.842°, 33.635°, 34.842°
$${another}\:\left({better}\right)\:{way}\:{to}\:{solve}: \\ $$$$ \\ $$$${the}\:{projection}\:{of}\:{the}\:{circle}\:{in}\:{the} \\ $$$${xy}−{plane}\:{is}\:{an}\:{ellipse}\:{which}\:{has}\:{the} \\ $$$${semi}−{major}\:{axis}\:{R}\:{and}\:{tangents} \\ $$$${x}−{axis}\:{and}\:{y}−{axis}\:{simultaneously}. \\ $$$$ \\ $$$${assume}\:{the}\:{semi}−{minor}\:{axis}\:{of}\:{ellipse} \\ $$$${is}\:{d}. \\ $$$${in}\:{coordinate}\:{system}\:{x}'{y}'\:{the}\:{eqn}. \\ $$$${of}\:{ellipse}\:{is} \\ $$$$\frac{{x}^{\mathrm{2}} }{{R}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{d}^{\mathrm{2}} }=\mathrm{1} \\ $$$${GA}=\frac{{k}}{\mathrm{sin}\:\alpha},\:{GB}=\frac{{h}}{\mathrm{cos}\:\alpha} \\ $$$$ \\ $$$${eqn}.\:{of}\:{line}\:{OA}\:{in}\:{x}'{y}'\:{system}\:{is} \\ $$$$\frac{{y}−{d}}{{x}−\frac{{k}}{\mathrm{sin}\:\alpha}}=\mathrm{tan}\:\alpha \\ $$$$\Rightarrow\mathrm{sin}\:\alpha\:{x}−\mathrm{cos}\:\alpha\:{y}+{d}\:\mathrm{cos}\:\alpha−{k}=\mathrm{0} \\ $$$${since}\:{it}'{s}\:{a}\:{tangent}\:{line}\:{of}\:{ellipse}, \\ $$$${R}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\alpha+{d}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\alpha−\left({d}\:\mathrm{cos}\:\alpha−{k}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{R}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\alpha−{k}^{\mathrm{2}} +\mathrm{2}{dk}\:\mathrm{cos}\:\alpha=\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${eqn}.\:{of}\:{line}\:{OB}\:{in}\:{x}'{y}'\:{system}\:{is} \\ $$$$\frac{{y}−{d}}{{x}+\frac{{h}}{\mathrm{cos}\:\alpha}}=−\frac{\mathrm{1}}{\mathrm{tan}\:\alpha} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha\:{x}+\mathrm{sin}\:\alpha\:{y}+{h}−{d}\:\mathrm{sin}\:\alpha=\mathrm{0} \\ $$$${since}\:{it}'{s}\:{a}\:{tangent}\:{line}\:{of}\:{ellipse}, \\ $$$${R}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\alpha+{q}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\alpha−\left({h}−{d}\:\mathrm{sin}\:\alpha\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{R}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\alpha−{h}^{\mathrm{2}} +\mathrm{2}{dh}\:\mathrm{sin}\:\alpha=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$${R}^{\mathrm{2}} −{k}^{\mathrm{2}} −{h}^{\mathrm{2}} +\mathrm{2}{d}\left({k}\:\mathrm{cos}\:\alpha+{h}\:\mathrm{sin}\:\alpha\right)=\mathrm{0} \\ $$$$\Rightarrow{d}=\frac{{k}^{\mathrm{2}} +{h}^{\mathrm{2}} −{R}^{\mathrm{2}} }{\mathrm{2}\left({k}\:\mathrm{cos}\:\alpha+{h}\:\mathrm{sin}\:\alpha\right)}\:\:\:…\left({iii}\right) \\ $$$${put}\:{this}\:{into}\:\left({ii}\right): \\ $$$${R}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\alpha+\frac{{h}\left({k}^{\mathrm{2}} +{h}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)\mathrm{sin}\:\alpha}{{k}\:\mathrm{cos}\:\alpha+{h}\:\mathrm{sin}\:\alpha}={h}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{{R}}{{h}}\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha}+\frac{\mathrm{1}+\left(\frac{{k}}{{h}}\right)^{\mathrm{2}} −\left(\frac{{R}}{{h}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{k}}{{h}}\right)\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}}=\mathrm{1}\:\:…\left({iv}\right) \\ $$$${or}\:{with}\:{t}=\mathrm{tan}\:\alpha,\:\lambda=\frac{{R}}{{h}},\:\mu=\frac{{k}}{{h}} \\ $$$$\Rightarrow\left(\mu^{\mathrm{2}} −\lambda^{\mathrm{2}} \right){t}^{\mathrm{3}} −\mu{t}^{\mathrm{2}} +\mu^{\mathrm{2}} {t}+\mu\left(\lambda^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0}\:\:\:…\left({iv}\right) \\ $$$$ \\ $$$${from}\:\left({iv}\right)\:{we}\:{can}\:{get}\:\alpha\:{at}\:{first}. \\ $$$${then}\:{we}\:{get}\:{d}\:{from}\:\left({iii}\right). \\ $$$$ \\ $$$$\mathrm{sin}\:\theta=\frac{{d}}{{R}}=\frac{{k}^{\mathrm{2}} +{h}^{\mathrm{2}} −{R}^{\mathrm{2}} }{\mathrm{2}{R}\left({k}\:\mathrm{cos}\:\alpha+{h}\:\mathrm{sin}\:\alpha\right)} \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}+\left(\frac{{k}}{{h}}\right)^{\mathrm{2}} −\left(\frac{{R}}{{h}}\right)^{\mathrm{2}} }{\mathrm{2}\left(\frac{{R}}{{h}}\right)\left[\mathrm{sin}\:\alpha+\left(\frac{{k}}{{h}}\right)\mathrm{cos}\:\:\alpha\right]} \\ $$$$ \\ $$$${examples}: \\ $$$${h}=\mathrm{4},\:{k}=\mathrm{3},\:{R}=\mathrm{4}: \\ $$$$\Rightarrow\alpha=\mathrm{0}°,\:\mathrm{29}.\mathrm{4346}° \\ $$$$\Rightarrow\theta=\mathrm{22}.\mathrm{024}°,\:\mathrm{14}.\mathrm{225}° \\ $$$$ \\ $$$${h}=\mathrm{3},\:{k}=\mathrm{3},\:{R}=\mathrm{2}.\mathrm{5}: \\ $$$$\Rightarrow\alpha=\mathrm{30}.\mathrm{8212}°,\:\mathrm{45}°,\:\mathrm{59}.\mathrm{1788}° \\ $$$$\Rightarrow\theta=\mathrm{34}.\mathrm{842}°,\:\mathrm{33}.\mathrm{635}°,\:\mathrm{34}.\mathrm{842}° \\ $$
Commented by ajfour last updated on 01/Nov/18
You did it Sir. This is quite  better. THANKS again.
$${You}\:{did}\:{it}\:{Sir}.\:{This}\:{is}\:{quite} \\ $$$${better}.\:\mathcal{THANKS}\:{again}. \\ $$
Commented by ajfour last updated on 02/Nov/18
I have posted an alternate way, Sir.
$${I}\:{have}\:{posted}\:{an}\:{alternate}\:{way},\:{Sir}. \\ $$

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