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Question Number 46308 by ajfour last updated on 23/Oct/18
Commented by ajfour last updated on 23/Oct/18
O is the centre of circle, C the centre of the unit square. If (i) α=tan^(−1) (4/3), (ii) α=60°, find radius R.
$${O}\:{is}\:{the}\:{centre}\:{of}\:{circle},\:{C}\:{the} \\ $$$${centre}\:{of}\:{the}\:{unit}\:{square}.\: \\ $$$${If}\:\left({i}\right)\:\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}},\:\:\left({ii}\right)\:\alpha=\mathrm{60}°,\: \\ $$$${find}\:{radius}\:{R}. \\ $$
Answered by MrW3 last updated on 23/Oct/18
Commented by MrW3 last updated on 23/Oct/18
let a=1 and b=(a/2) OQ=R−b CQ=b OC=(√(b^2 +(R−b)^2 )) OT=2b−R OP=R PT=(√(R^2 −(2b−R)^2 )) CQ−PT=b−(√(R^2 −(2b−R)^2 )) CP^2 =b^2 +[b−(√(R^2 −(2b−R)^2 ))]^2 =2b[2R−(√(4b(R−b)))−b] R^2 =b^2 +(R−b)^2 +2b[2R−(√(4b(R−b)))−b]−2(√(2b[b^2 +(R−b)^2 ][2R−(√(4b(R−b)))−b])) cos α 0=2b(R+b)−2b(√(4b(R−b)))−2b^2 −2(√(2b[b^2 +(R−b)^2 ][2R−(√(4b(R−b)))−b])) cos α let λ=(R/b)=((2R)/a)=2R λ−2(√(λ−1))−(√(2[1+(λ−1)^2 ][2λ−2(√(λ−1))−1])) cos α=0 with α=60° 2λ−4(√(λ−1))−(√(2[1+(λ−1)^2 ][2λ−2(√(λ−1))−1]))=0 ⇒λ≈1.0636 ⇒R≈0.5318 with α=tan^(−1) (4/3) ⇒λ≈1.0196 ⇒R≈0.5098
$${let}\:{a}=\mathrm{1}\:{and}\:{b}=\frac{{a}}{\mathrm{2}} \\ $$$${OQ}={R}−{b} \\ $$$${CQ}={b} \\ $$$${OC}=\sqrt{{b}^{\mathrm{2}} +\left({R}−{b}\right)^{\mathrm{2}} } \\ $$$${OT}=\mathrm{2}{b}−{R} \\ $$$${OP}={R} \\ $$$${PT}=\sqrt{{R}^{\mathrm{2}} −\left(\mathrm{2}{b}−{R}\right)^{\mathrm{2}} } \\ $$$${CQ}−{PT}={b}−\sqrt{{R}^{\mathrm{2}} −\left(\mathrm{2}{b}−{R}\right)^{\mathrm{2}} } \\ $$$${CP}^{\mathrm{2}} ={b}^{\mathrm{2}} +\left[{b}−\sqrt{{R}^{\mathrm{2}} −\left(\mathrm{2}{b}−{R}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} \\ $$$$=\mathrm{2}{b}\left[\mathrm{2}{R}−\sqrt{\mathrm{4b}\left(\mathrm{R}−{b}\right)}−{b}\right] \\ $$$$ \\ $$$${R}^{\mathrm{2}} ={b}^{\mathrm{2}} +\left({R}−{b}\right)^{\mathrm{2}} +\mathrm{2}{b}\left[\mathrm{2}{R}−\sqrt{\mathrm{4b}\left(\mathrm{R}−{b}\right)}−{b}\right]−\mathrm{2}\sqrt{\mathrm{2}{b}\left[{b}^{\mathrm{2}} +\left({R}−{b}\right)^{\mathrm{2}} \right]\left[\mathrm{2}{R}−\sqrt{\mathrm{4b}\left(\mathrm{R}−{b}\right)}−{b}\right]}\:\mathrm{cos}\:\alpha \\ $$$$\mathrm{0}=\mathrm{2}{b}\left({R}+{b}\right)−\mathrm{2}{b}\sqrt{\mathrm{4b}\left(\mathrm{R}−{b}\right)}−\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}{b}\left[{b}^{\mathrm{2}} +\left({R}−{b}\right)^{\mathrm{2}} \right]\left[\mathrm{2}{R}−\sqrt{\mathrm{4b}\left(\mathrm{R}−{b}\right)}−{b}\right]}\:\mathrm{cos}\:\alpha \\ $$$${let}\:\lambda=\frac{{R}}{{b}}=\frac{\mathrm{2}{R}}{{a}}=\mathrm{2}{R} \\ $$$$\lambda−\mathrm{2}\sqrt{\lambda−\mathrm{1}}−\sqrt{\mathrm{2}\left[\mathrm{1}+\left(\lambda−\mathrm{1}\right)^{\mathrm{2}} \right]\left[\mathrm{2}\lambda−\mathrm{2}\sqrt{\lambda−\mathrm{1}}−\mathrm{1}\right]}\:\mathrm{cos}\:\alpha=\mathrm{0} \\ $$$${with}\:\alpha=\mathrm{60}° \\ $$$$\mathrm{2}\lambda−\mathrm{4}\sqrt{\lambda−\mathrm{1}}−\sqrt{\mathrm{2}\left[\mathrm{1}+\left(\lambda−\mathrm{1}\right)^{\mathrm{2}} \right]\left[\mathrm{2}\lambda−\mathrm{2}\sqrt{\lambda−\mathrm{1}}−\mathrm{1}\right]}=\mathrm{0} \\ $$$$\Rightarrow\lambda\approx\mathrm{1}.\mathrm{0636} \\ $$$$\Rightarrow{R}\approx\mathrm{0}.\mathrm{5318} \\ $$$$ \\ $$$${with}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow\lambda\approx\mathrm{1}.\mathrm{0196} \\ $$$$\Rightarrow{R}\approx\mathrm{0}.\mathrm{5098} \\ $$
Commented by ajfour last updated on 23/Oct/18
correct answer sir, thank you!
$${correct}\:{answer}\:{sir},\:{thank}\:{you}! \\ $$
Answered by ajfour last updated on 23/Oct/18
Commented by ajfour last updated on 23/Oct/18
α = θ+φ tan θ = ((a/2)/(R−(a/2))) let λ=(R/a) ⇒ tan θ = (1/(2λ−1)) =(1/t^2 ) (say) tan φ= ((PM)/(CM)) = (((√(R^2 −(a−R)^2 ))−(a/2))/(a/2)) = 2(√(2λ−1))−1 = 2t−1 let m=tan α ⇒ m=tan (θ+φ) m−mtan θ tan φ = tan θ+tan φ m−((m(2t−1))/t^2 ) = (1/t^2 )+2t−1 ⇒ mt^2 −2mt+m=1+2t^3 −t^2 2t^3 −(1+m)t^2 +2mt−(m−1)=0 for α=tan^(−1) (4/3) , m=(4/3) 2t^3 −((7t^2 )/3)+((8t)/3)−(1/3)=0 or 6t^3 −7t^2 +8t−1 = 0 t ≈ 0.14012 λ =(R/a)= ((t^2 +1)/2) R ≈ 0.5098 .
$$\alpha\:=\:\theta+\phi \\ $$$$\mathrm{tan}\:\theta\:=\:\frac{{a}/\mathrm{2}}{{R}−\frac{{a}}{\mathrm{2}}} \\ $$$${let}\:\:\lambda=\frac{{R}}{{a}}\: \\ $$$$\:\Rightarrow\:\mathrm{tan}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}\lambda−\mathrm{1}}\:=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\:\:\left({say}\right) \\ $$$$\mathrm{tan}\:\phi=\:\frac{{PM}}{{CM}}\:=\:\frac{\sqrt{{R}^{\mathrm{2}} −\left({a}−{R}\right)^{\mathrm{2}} }−\frac{{a}}{\mathrm{2}}}{{a}/\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\sqrt{\mathrm{2}\lambda−\mathrm{1}}−\mathrm{1}\:=\:\mathrm{2}{t}−\mathrm{1} \\ $$$${let}\:\:{m}=\mathrm{tan}\:\alpha \\ $$$$\Rightarrow\:\:{m}=\mathrm{tan}\:\left(\theta+\phi\right) \\ $$$${m}−{m}\mathrm{tan}\:\theta\:\mathrm{tan}\:\phi\:=\:\mathrm{tan}\:\theta+\mathrm{tan}\:\phi \\ $$$${m}−\frac{{m}\left(\mathrm{2}{t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{2}{t}−\mathrm{1} \\ $$$$\Rightarrow\:\:{mt}^{\mathrm{2}} −\mathrm{2}{mt}+{m}=\mathrm{1}+\mathrm{2}{t}^{\mathrm{3}} −{t}^{\mathrm{2}} \\ $$$$\mathrm{2}\boldsymbol{{t}}^{\mathrm{3}} −\left(\mathrm{1}+\boldsymbol{{m}}\right)\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{mt}}−\left(\boldsymbol{{m}}−\mathrm{1}\right)=\mathrm{0} \\ $$$${for}\:\:\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}\:\:\:,\:\:{m}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{2}{t}^{\mathrm{3}} −\frac{\mathrm{7}{t}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{8}{t}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{0} \\ $$$${or}\:\:\:\:\mathrm{6}\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{7}\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{t}}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:\:\:\boldsymbol{{t}}\:\approx\:\mathrm{0}.\mathrm{14012} \\ $$$$\:\:\lambda\:=\frac{{R}}{{a}}=\:\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}}\: \\ $$$$\:\:\:{R}\:\approx\:\mathrm{0}.\mathrm{5098}\:. \\ $$

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