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Question Number 189304 by ajfour last updated on 14/Mar/23
Commented by ajfour last updated on 14/Mar/23
Find largest radius ball beneath cone and wirhin cyclinder.
$${Find}\:{largest}\:{radius}\:{ball}\:{beneath} \\ $$$${cone}\:{and}\:{wirhin}\:{cyclinder}. \\ $$
Answered by mr W last updated on 14/Mar/23
Commented by mr W last updated on 14/Mar/23
Commented by mr W last updated on 15/Mar/23
tan β=(b/(2a)) m=tan (β+θ)=(((b/(2a))+tan θ)/(1−((b tan θ)/(2a))))=((b+2a tan θ)/(2a−b tan θ)) center of ball: C[(a−r)cos φ, (a−r)sin φ, r] vertex of cone: B(−a,0,0) axis of cone: BG^(→) =(1, 0, m) BC^(→) =((a−r)cos φ+a, (a−r)sin φ, r) BC=(√([(a−r)cos φ+a]^2 +(a−r)^2 sin^2 φ+r^2 )) BC=(√(a^2 +(a−r)^2 +r^2 +2a(a−r)cos φ)) cos (α+θ)=((BG^(→) ∙BC^(→) )/(∣BG^(→) ∣∣BC^(→) ∣))=(((a−r)cos φ+a+mr)/( BC×(√(1+m^2 )))) BC cos α cos θ−BC sin α sin θ=(((a−r)cos φ+a+mr)/( (√(1+m^2 )))) cos θ (√(a^2 +(a−r)^2 +2a(a−r)cos φ))−r sin θ=(((a−r)cos φ+a+mr)/( (√(1+m^2 )))) let λ=(r/a) ⇒cos θ (√(1+(1−λ)^2 +2(1−λ) cos φ))−λ sin θ=(((1−λ) cos φ+1+mλ)/( (√(1+m^2 )))) or [(1+m^2 )cos^2 θ−(m+sin θ (√(1+m^2 ))−cos φ)^2 ]λ^2 −2(1+cos φ)[(1+m^2 )cos^2 θ+m+sin θ (√(1+m^2 ))−cos φ]λ+2(1+cos φ)(1+m^2 )cos^2 θ−(1+cos φ)^2 =0 ⇒λ=(((1+cos φ)[(1+m^2 )cos^2 θ+m+sin θ (√(1+m^2 ))−cos φ]−(√((1+cos φ)^2 [(1+m^2 )cos^2 θ+m+sin θ (√(1+m^2 ))−cos φ]^2 −[(1+m^2 )cos^2 θ−(m+sin θ (√(1+m^2 ))−cos φ)^2 ][2(1+cos φ)(1+m^2 )cos^2 θ−(1+cos φ)^2 ])))/((1+m^2 )cos^2 θ−(m+sin θ (√(1+m^2 ))−cos φ)^2 )) λ_(max) is not at φ=0° as following example shows. example: a=1, b=1, θ=30° ⇒r_(max) ≈0.3823 at φ≈28.1826°
$$\mathrm{tan}\:\beta=\frac{{b}}{\mathrm{2}{a}} \\ $$$${m}=\mathrm{tan}\:\left(\beta+\theta\right)=\frac{\frac{{b}}{\mathrm{2}{a}}+\mathrm{tan}\:\theta}{\mathrm{1}−\frac{{b}\:\mathrm{tan}\:\theta}{\mathrm{2}{a}}}=\frac{{b}+\mathrm{2}{a}\:\mathrm{tan}\:\theta}{\mathrm{2}{a}−{b}\:\mathrm{tan}\:\theta} \\ $$$${center}\:{of}\:{ball}: \\ $$$${C}\left[\left({a}−{r}\right)\mathrm{cos}\:\phi,\:\left({a}−{r}\right)\mathrm{sin}\:\phi,\:{r}\right] \\ $$$${vertex}\:{of}\:{cone}: \\ $$$${B}\left(−{a},\mathrm{0},\mathrm{0}\right) \\ $$$${axis}\:{of}\:{cone}: \\ $$$$\overset{\rightarrow} {{BG}}=\left(\mathrm{1},\:\mathrm{0},\:{m}\right) \\ $$$$\overset{\rightarrow} {{BC}}=\left(\left({a}−{r}\right)\mathrm{cos}\:\phi+{a},\:\:\left({a}−{r}\right)\mathrm{sin}\:\phi,\:{r}\right) \\ $$$${BC}=\sqrt{\left[\left({a}−{r}\right)\mathrm{cos}\:\phi+{a}\right]^{\mathrm{2}} +\left({a}−{r}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\phi+{r}^{\mathrm{2}} } \\ $$$${BC}=\sqrt{{a}^{\mathrm{2}} +\left({a}−{r}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{a}\left({a}−{r}\right)\mathrm{cos}\:\phi} \\ $$$$\mathrm{cos}\:\left(\alpha+\theta\right)=\frac{\overset{\rightarrow} {{BG}}\centerdot\overset{\rightarrow} {{BC}}}{\mid\overset{\rightarrow} {{BG}}\mid\mid\overset{\rightarrow} {{BC}}\mid}=\frac{\left({a}−{r}\right)\mathrm{cos}\:\phi+{a}+{mr}}{\:{BC}×\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }} \\ $$$${BC}\:\mathrm{cos}\:\alpha\:\mathrm{cos}\:\theta−{BC}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\theta=\frac{\left({a}−{r}\right)\mathrm{cos}\:\phi+{a}+{mr}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\theta\:\sqrt{{a}^{\mathrm{2}} +\left({a}−{r}\right)^{\mathrm{2}} +\mathrm{2}{a}\left({a}−{r}\right)\mathrm{cos}\:\phi}−{r}\:\mathrm{sin}\:\theta=\frac{\left({a}−{r}\right)\mathrm{cos}\:\phi+{a}+{mr}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }} \\ $$$${let}\:\lambda=\frac{{r}}{{a}} \\ $$$$\Rightarrow\mathrm{cos}\:\theta\:\sqrt{\mathrm{1}+\left(\mathrm{1}−\lambda\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}−\lambda\right)\:\mathrm{cos}\:\phi}−\lambda\:\mathrm{sin}\:\theta=\frac{\left(\mathrm{1}−\lambda\right)\:\mathrm{cos}\:\phi+\mathrm{1}+{m}\lambda}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }} \\ $$$${or} \\ $$$$\left[\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta−\left({m}+\mathrm{sin}\:\theta\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }−\mathrm{cos}\:\phi\right)^{\mathrm{2}} \right]\lambda^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left[\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta+{m}+\mathrm{sin}\:\theta\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }−\mathrm{cos}\:\phi\right]\lambda+\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta−\left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left[\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta+{m}+\mathrm{sin}\:\theta\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }−\mathrm{cos}\:\phi\right]−\sqrt{\left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} \left[\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta+{m}+\mathrm{sin}\:\theta\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }−\mathrm{cos}\:\phi\right]^{\mathrm{2}} −\left[\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta−\left({m}+\mathrm{sin}\:\theta\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }−\mathrm{cos}\:\phi\right)^{\mathrm{2}} \right]\left[\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta−\left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} \right]}}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta−\left({m}+\mathrm{sin}\:\theta\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }−\mathrm{cos}\:\phi\right)^{\mathrm{2}} } \\ $$$$\lambda_{{max}} \:{is}\:{not}\:{at}\:\phi=\mathrm{0}°\:{as}\:{following} \\ $$$${example}\:{shows}. \\ $$$${example}:\:{a}=\mathrm{1},\:{b}=\mathrm{1},\:\theta=\mathrm{30}°\: \\ $$$$\Rightarrow{r}_{{max}} \approx\mathrm{0}.\mathrm{3823}\:{at}\:\phi\approx\mathrm{28}.\mathrm{1826}° \\ $$
Commented by mr W last updated on 14/Mar/23
Commented by ajfour last updated on 23/Mar/23
Sir i had gotten busy, so i posted my solution after a week and thank you incredible effort you have put in. I truly believe your answer should be right, but can u cast a glance and help in finding mistake in mine ..
Commented by mr W last updated on 12/Apr/23
sorry i saw this message right now, because i don′t get notification automatically. i′ll try to follow your solution.
$${sorry}\:{i}\:{saw}\:{this}\:{message}\:{right}\:{now}, \\ $$$${because}\:{i}\:{don}'{t}\:{get}\:{notification} \\ $$$${automatically}. \\ $$$${i}'{ll}\:{try}\:{to}\:{follow}\:{your}\:{solution}. \\ $$
Answered by ajfour last updated on 22/Mar/23
Commented by ajfour last updated on 23/Mar/23
tan θ=m tan β=(b/(2a)) tan (θ+β)=q=((m+(b/(2a)))/(1−((bm)/(2a))))=((2am+b)/(2a−mb)) AJ=AD=s BJ=ms BH=r+ms DP=(r+ms)cos δ ABsin (θ+β)=r+(r+ms)sin δ AB=(s/(cos θ))=s(√(1+m^2 )) ⇒ ((ssin (θ+β))/(cos θ))=r+(r+ms)sin δ ......(i) let CP=p=a−AP AP=ABcos (θ+β) ⇒ p=a−((scos (θ+β))/(cos θ)) cos φ=((DP^2 −(a−r)^2 −p^2 )/(2p(a−r))) also cos φ=((s^2 −(a−r)^2 −a^2 )/(2a(a−r))) let (r/a)=x , (s/a)=y , (p/a)=z ⇒ (x+my)^2 cos^2 δ−(1−x)^2 −z^2 =z{y^2 −(1−x)^2 −1} ....(ii) p=a−((scos (θ+β))/(cos θ)) ⇒ z=1−((ycos (θ+β))/(cos θ)) ((ssin (θ+β))/(cos θ))=r+(r+ms)sin δ ⇒ (x+my)^2 sin^2 δ={((ysin (θ+β))/(cos θ))−x}^2 Adding to (ii) (x+my)^2 = (1−x)^2 +z^2 +z{y^2 −(1−x)^2 −1} +{((ysin (θ+β))/(cos θ))−x}^2 now z=1−((ycos (θ+β))/(cos θ)) ⇒ (x+((ysin θ)/(cos θ)))^2 = (1−x)^2 +{1−((ycos (θ+β))/(cos θ))}^2 +{1−((ycos (θ+β))/(cos θ))}{y^2 −(1−x)^2 −1} +{((ysin (θ+β))/(cos θ))−x}^2 ⇒ x^2 +((y^2 sin^2 θ)/(cos^2 θ))+((2xysin θ)/(cos θ))=(1−x)^2 +1 +((y^2 cos^2 (θ+β))/(cos^2 θ))−((2ycos (θ+β))/(cos θ)) +y^2 −1−(1−x)^2 +x^2 +((y^2 sin^2 (θ+β))/(cos^2 θ))−((2xysin (θ+β))/(cos θ)) +{1+(1−x)^2 −y^2 }((ycos (θ+β))/(cos θ)) ⇒ ((2xy)/(cos θ)){sin θ+sin (θ+β)} =2y^2 +((y(x^2 −2x−y^2 )cos (θ+β))/(cos θ)) ⇒ 2x{sin θ+sin (θ+β)} =2ycos θ+(x^2 −2x−y^2 )cos (θ+β) differentiating w.r.t. y with (dx/dy)=0 ⇒ 2cos θ=2ycos (θ+β) ⇒ y=((cos θ)/(cos (θ+β))) ⇒ 2x_m {sin θ+sin (θ+β)}cos (θ+β) =x_m (x_m −2)cos^2 (θ+β)+cos^2 θ If θ=30° tan β=(b/(2a))=(1/2) , then tan (θ+β)=((2+(√3))/(2(√3)−1)) hence 2x_m {(1/2)+((2+(√3))/(2(√5)))}(((2(√3)−1))/(2(√5))) =x_m (x_m −2)(((2(√3)−1)^2 )/(20))+(3/4) ⇒ 2x_m {((2(√3)−1)/(4(√5)))+(((2(√3)−1)^2 )/(10))} =(((2(√3)−1)^2 x_m ^2 )/(20))+(3/4) ⇒ x_m ^2 −2x_m {((√5)/(2(√3)−1))+2}+((15)/((2(√3)−1)^2 ))=0 say (2(√3)−1)x_m =q q^2 −2q((√5)+4(√3)−2)+15=0 ⇒ q ≈ 1.1371 x_m =(r_m /a)=(q/((2(√3)−1)))≈ 0.4615
$$\mathrm{tan}\:\theta={m} \\ $$$$\mathrm{tan}\:\beta=\frac{{b}}{\mathrm{2}{a}} \\ $$$$\mathrm{tan}\:\left(\theta+\beta\right)={q}=\frac{{m}+\frac{{b}}{\mathrm{2}{a}}}{\mathrm{1}−\frac{{bm}}{\mathrm{2}{a}}}=\frac{\mathrm{2}{am}+{b}}{\mathrm{2}{a}−{mb}} \\ $$$${AJ}={AD}={s}\:\:\:\:{BJ}={ms} \\ $$$${BH}={r}+{ms} \\ $$$${DP}=\left({r}+{ms}\right)\mathrm{cos}\:\delta \\ $$$${AB}\mathrm{sin}\:\left(\theta+\beta\right)={r}+\left({r}+{ms}\right)\mathrm{sin}\:\delta \\ $$$$\:\:\:\:{AB}=\frac{{s}}{\mathrm{cos}\:\theta}={s}\sqrt{\mathrm{1}+{m}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\frac{{s}\mathrm{sin}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta}={r}+\left({r}+{ms}\right)\mathrm{sin}\:\delta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:……\left({i}\right) \\ $$$${let}\:\:{CP}={p}={a}−{AP} \\ $$$${AP}={AB}\mathrm{cos}\:\left(\theta+\beta\right) \\ $$$$\Rightarrow\:\:{p}={a}−\frac{{s}\mathrm{cos}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta} \\ $$$$\mathrm{cos}\:\phi=\frac{{DP}\:^{\mathrm{2}} −\left({a}−{r}\right)^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{2}{p}\left({a}−{r}\right)} \\ $$$${also} \\ $$$$\mathrm{cos}\:\phi=\frac{{s}^{\mathrm{2}} −\left({a}−{r}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{a}\left({a}−{r}\right)} \\ $$$${let}\:\:\frac{{r}}{{a}}={x}\:\:,\:\:\:\frac{{s}}{{a}}={y}\:\:,\:\:\frac{{p}}{{a}}={z} \\ $$$$\Rightarrow\:\:\left({x}+{my}\right)^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \delta−\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −{z}^{\mathrm{2}} \\ $$$$\:={z}\left\{{y}^{\mathrm{2}} −\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{1}\right\}\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$$\:\:{p}={a}−\frac{{s}\mathrm{cos}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\:{z}=\mathrm{1}−\frac{{y}\mathrm{cos}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta} \\ $$$$\:\:\frac{{s}\mathrm{sin}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta}={r}+\left({r}+{ms}\right)\mathrm{sin}\:\delta \\ $$$$\Rightarrow \\ $$$$\:\left({x}+{my}\right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \delta=\left\{\frac{{y}\mathrm{sin}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta}−{x}\right\}^{\mathrm{2}} \\ $$$${Adding}\:{to}\:\:\left({ii}\right) \\ $$$$\left({x}+{my}\right)^{\mathrm{2}} = \\ $$$$\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +{z}^{\mathrm{2}} +{z}\left\{{y}^{\mathrm{2}} −\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{1}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left\{\frac{{y}\mathrm{sin}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta}−{x}\right\}^{\mathrm{2}} \\ $$$${now}\:\:\:{z}=\mathrm{1}−\frac{{y}\mathrm{cos}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta} \\ $$$$\Rightarrow \\ $$$$\:\:\left({x}+\frac{{y}\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right)^{\mathrm{2}} = \\ $$$$\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +\left\{\mathrm{1}−\frac{{y}\mathrm{cos}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta}\right\}^{\mathrm{2}} \\ $$$$\:+\left\{\mathrm{1}−\frac{{y}\mathrm{cos}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta}\right\}\left\{{y}^{\mathrm{2}} −\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{1}\right\} \\ $$$$\:\:\:\:\:\:+\left\{\frac{{y}\mathrm{sin}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta}−{x}\right\}^{\mathrm{2}} \\ $$$$\Rightarrow\: \\ $$$$\cancel{{x}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{cos}\:^{\mathrm{2}} \theta}+\frac{\mathrm{2}{xy}\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}=\cancel{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\cancel{\mathrm{1}} \\ $$$$+\frac{{y}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \left(\theta+\beta\right)}{\mathrm{cos}\:^{\mathrm{2}} \theta}−\frac{\mathrm{2}{y}\mathrm{cos}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta} \\ $$$$+{y}^{\mathrm{2}} −\cancel{\mathrm{1}}−\cancel{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$+\cancel{{x}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \left(\theta+\beta\right)}{\mathrm{cos}\:^{\mathrm{2}} \theta}−\frac{\mathrm{2}{xy}\mathrm{sin}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta} \\ $$$$+\left\{\mathrm{1}+\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −{y}^{\mathrm{2}} \right\}\frac{{y}\mathrm{cos}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta} \\ $$$$\Rightarrow \\ $$$$\frac{\mathrm{2}{xy}}{\mathrm{cos}\:\theta}\left\{\mathrm{sin}\:\theta+\mathrm{sin}\:\left(\theta+\beta\right)\right\} \\ $$$$=\mathrm{2}{y}^{\mathrm{2}} +\frac{{y}\left({x}^{\mathrm{2}} −\mathrm{2}{x}−{y}^{\mathrm{2}} \right)\mathrm{cos}\:\left(\theta+\beta\right)}{\mathrm{cos}\:\theta} \\ $$$$\Rightarrow \\ $$$$\:\:\mathrm{2}{x}\left\{\mathrm{sin}\:\theta+\mathrm{sin}\:\left(\theta+\beta\right)\right\} \\ $$$$\:\:\:=\mathrm{2}{y}\mathrm{cos}\:\theta+\left({x}^{\mathrm{2}} −\mathrm{2}{x}−{y}^{\mathrm{2}} \right)\mathrm{cos}\:\left(\theta+\beta\right) \\ $$$${differentiating}\:{w}.{r}.{t}.\:\:\:\:{y}\:\:\:{with} \\ $$$$\frac{{dx}}{{dy}}=\mathrm{0}\:\:\:\Rightarrow \\ $$$$\mathrm{2cos}\:\theta=\mathrm{2}{y}\mathrm{cos}\:\left(\theta+\beta\right) \\ $$$$\Rightarrow\:\:{y}=\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\theta+\beta\right)} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}{x}_{{m}} \left\{\mathrm{sin}\:\theta+\mathrm{sin}\:\left(\theta+\beta\right)\right\}\mathrm{cos}\:\left(\theta+\beta\right) \\ $$$$\:\:={x}_{{m}} \left({x}_{{m}} −\mathrm{2}\right)\mathrm{cos}\:^{\mathrm{2}} \left(\theta+\beta\right)+\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$${If}\:\:\theta=\mathrm{30}°\:\:\:\mathrm{tan}\:\beta=\frac{{b}}{\mathrm{2}{a}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:,\:\:{then} \\ $$$$\mathrm{tan}\:\left(\theta+\beta\right)=\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}\:\:\:\: \\ $$$${hence} \\ $$$$\mathrm{2}{x}_{{m}} \left\{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{5}}}\right\}\frac{\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\:\:\:={x}_{{m}} \left({x}_{{m}} −\mathrm{2}\right)\frac{\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{20}}+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{2}{x}_{{m}} \left\{\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{5}}}+\frac{\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{10}}\right\} \\ $$$$\:\:=\frac{\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} {x}_{{m}} ^{\mathrm{2}} }{\mathrm{20}}+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow \\ $$$$\:{x}_{{m}} ^{\mathrm{2}} −\mathrm{2}{x}_{{m}} \left\{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}+\mathrm{2}\right\}+\frac{\mathrm{15}}{\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$${say}\:\:\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right){x}_{{m}} ={q} \\ $$$${q}^{\mathrm{2}} −\mathrm{2}{q}\left(\sqrt{\mathrm{5}}+\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{2}\right)+\mathrm{15}=\mathrm{0} \\ $$$$\Rightarrow\:\:{q}\:\approx\:\mathrm{1}.\mathrm{1371} \\ $$$${x}_{{m}} =\frac{{r}_{{m}} }{{a}}=\frac{{q}}{\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)}\approx\:\mathrm{0}.\mathrm{4615} \\ $$$$ \\ $$

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