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Question Number 72952 by ajfour last updated on 05/Nov/19
Commented by ajfour last updated on 05/Nov/19
AB divides the composite figure of square mounted by a semicircle in two equal parts. Determine α.
$${AB}\:{divides}\:{the}\:{composite}\:{figure} \\ $$$${of}\:{square}\:{mounted}\:{by}\:{a}\:{semicircle} \\ $$$${in}\:{two}\:{equal}\:{parts}.\:{Determine}\:\alpha. \\ $$
Commented by mind is power last updated on 05/Nov/19
let O center of semi circl,β angl between OB (π/2)R^2 +4R^2 =(R^2 /2).(β/)−(1/2)(R.sin(β).(Rcos(β)−((Rsin(β))/(tg(α)))))+{(R−(Rcosβ−((Rsin(β))/(tg(α)))))+2R}.R ⇔(π/2)+4=(β/2) −(1/2)(sin(β)cos(β)−((sin^2 (β))/(tg(α))))+(3−cos(β)+((sin(β))/(tg(α)))) ⇒x=(π/2)+4 ⇒x−(β/2)=((1/(tg(α))))(((sin^2 (β))/2)+sin(β))−((sin(β)cos(β))/2)+3−cos(β) (1/(tg(α)))=((x−(β/2)+((sin(β)cos(β))/2)+cos(β)−3)/(((sin^2 (β))/2)+sin(β))) ((2R)/(tg(α)))−R=R+Rcos(β)−((Rsin(β))/(tg(α))) (1/(tg(α)))(2+sin(β))=((2+cos(β))/) (1/(tg(α)))=((2+cos(β))/(2+sin(β)))=((x−(β/2)+((sin(β)cos(β))/2)+cos(β)−3)/(((sin^2 (β))/2)+sin(β))) ⇒sin^2 (β)+2sin(β)=−3sin(β)−6+2x−(β/2)(2+sin(β))+sin(β)cos(β)) let β_0 of this equation ⇒α=cot^(−1) (((x−(β_0 /2)+((sin(β_0 )cos(β_0 ))/2)+cos(β_0 )−3)/(((sin^2 (β_0 ))/2)+sin(β_0 ))))
$$\mathrm{let}\:\mathrm{O}\:\mathrm{center}\:\mathrm{of}\:\mathrm{semi}\:\mathrm{circl},\beta\:\mathrm{angl}\:\mathrm{between}\:\mathrm{OB} \\ $$$$\frac{\pi}{\mathrm{2}}\mathrm{R}^{\mathrm{2}} +\mathrm{4R}^{\mathrm{2}} =\frac{\mathrm{R}^{\mathrm{2}} }{\mathrm{2}}.\frac{\beta}{}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{R}.\mathrm{sin}\left(\beta\right).\left(\mathrm{Rcos}\left(\beta\right)−\frac{\mathrm{Rsin}\left(\beta\right)}{\mathrm{tg}\left(\alpha\right)}\right)\right)+\left\{\left(\mathrm{R}−\left(\mathrm{Rcos}\beta−\frac{\mathrm{Rsin}\left(\beta\right)}{\mathrm{tg}\left(\alpha\right)}\right)\right)+\mathrm{2R}\right\}.\mathrm{R} \\ $$$$\Leftrightarrow\frac{\pi}{\mathrm{2}}+\mathrm{4}=\frac{\beta}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\left(\beta\right)\mathrm{cos}\left(\beta\right)−\frac{\mathrm{sin}^{\mathrm{2}} \left(\beta\right)}{\mathrm{tg}\left(\alpha\right)}\right)+\left(\mathrm{3}−\mathrm{cos}\left(\beta\right)+\frac{\mathrm{sin}\left(\beta\right)}{\mathrm{tg}\left(\alpha\right)}\right) \\ $$$$\Rightarrow\mathrm{x}=\frac{\pi}{\mathrm{2}}+\mathrm{4} \\ $$$$\Rightarrow\mathrm{x}−\frac{\beta}{\mathrm{2}}=\left(\frac{\mathrm{1}}{\mathrm{tg}\left(\alpha\right)}\right)\left(\frac{\mathrm{sin}^{\mathrm{2}} \left(\beta\right)}{\mathrm{2}}+\mathrm{sin}\left(\beta\right)\right)−\frac{\mathrm{sin}\left(\beta\right)\mathrm{cos}\left(\beta\right)}{\mathrm{2}}+\mathrm{3}−\mathrm{cos}\left(\beta\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{tg}\left(\alpha\right)}=\frac{\mathrm{x}−\frac{\beta}{\mathrm{2}}+\frac{\mathrm{sin}\left(\beta\right)\mathrm{cos}\left(\beta\right)}{\mathrm{2}}+\mathrm{cos}\left(\beta\right)−\mathrm{3}}{\frac{\mathrm{sin}^{\mathrm{2}} \left(\beta\right)}{\mathrm{2}}+\mathrm{sin}\left(\beta\right)} \\ $$$$\frac{\mathrm{2R}}{\mathrm{tg}\left(\alpha\right)}−\mathrm{R}=\mathrm{R}+\mathrm{Rcos}\left(\beta\right)−\frac{\mathrm{Rsin}\left(\beta\right)}{\mathrm{tg}\left(\alpha\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{tg}\left(\alpha\right)}\left(\mathrm{2}+\mathrm{sin}\left(\beta\right)\right)=\frac{\mathrm{2}+\mathrm{cos}\left(\beta\right)}{} \\ $$$$\frac{\mathrm{1}}{\mathrm{tg}\left(\alpha\right)}=\frac{\mathrm{2}+\mathrm{cos}\left(\beta\right)}{\mathrm{2}+\mathrm{sin}\left(\beta\right)}=\frac{\mathrm{x}−\frac{\beta}{\mathrm{2}}+\frac{\mathrm{sin}\left(\beta\right)\mathrm{cos}\left(\beta\right)}{\mathrm{2}}+\mathrm{cos}\left(\beta\right)−\mathrm{3}}{\frac{\mathrm{sin}^{\mathrm{2}} \left(\beta\right)}{\mathrm{2}}+\mathrm{sin}\left(\beta\right)} \\ $$$$\left.\Rightarrow\mathrm{sin}^{\mathrm{2}} \left(\beta\right)+\mathrm{2sin}\left(\beta\right)=−\mathrm{3sin}\left(\beta\right)−\mathrm{6}+\mathrm{2x}−\frac{\beta}{\mathrm{2}}\left(\mathrm{2}+\mathrm{sin}\left(\beta\right)\right)+\mathrm{sin}\left(\beta\right)\mathrm{cos}\left(\beta\right)\right) \\ $$$$\mathrm{let}\:\beta_{\mathrm{0}} \:\mathrm{of}\:\mathrm{this}\:\mathrm{equation} \\ $$$$\Rightarrow\alpha=\mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{x}−\frac{\beta_{\mathrm{0}} }{\mathrm{2}}+\frac{\mathrm{sin}\left(\beta_{\mathrm{0}} \right)\mathrm{cos}\left(\beta_{\mathrm{0}} \right)}{\mathrm{2}}+\mathrm{cos}\left(\beta_{\mathrm{0}} \right)−\mathrm{3}}{\frac{\mathrm{sin}^{\mathrm{2}} \left(\beta_{\mathrm{0}} \right)}{\mathrm{2}}+\mathrm{sin}\left(\beta_{\mathrm{0}} \right)}\right)\:\: \\ $$
Commented by MJS last updated on 05/Nov/19
I found no exact solution α≈54.8872°
$$\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\mathrm{exact}\:\mathrm{solution} \\ $$$$\alpha\approx\mathrm{54}.\mathrm{8872}° \\ $$
Commented by ajfour last updated on 05/Nov/19
Thanks Powerful Mind, and MjS Sir.
$${Thanks}\:{Powerful}\:{Mind},\:{and}\:{MjS} \\ $$$${Sir}. \\ $$
Answered by ajfour last updated on 05/Nov/19
Commented by ajfour last updated on 05/Nov/19
Area(sector BCF)=(β/2) tan α=((2+sin β)/(1+cos β)) ⇒ cot α=((1+cos β)/(2+sin β)) (1/2)(total area)=Area(right half of composite area) (π/4)+1=cot α+2(2−2cot α)+(β/2) −(1/2)×2(cot α−1)sin β say cot α=t (π/4)+1=t+4−4t+(β/2)−(t−1)sin β ⇒ (π/4)+1=4+(β/2)+sin β−(3+sin β)t ⇒ (3+sin β)(((1+cos β)/(2+sin β)))−(β/2)−sin β =3−(π/4) this eq. yields β and then α=tan^(−1) (((2+sin β)/(1+cos β))) .
$${Area}\left({sector}\:{BCF}\right)=\frac{\beta}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}+\mathrm{sin}\:\beta}{\mathrm{1}+\mathrm{cos}\:\beta}\:\:\Rightarrow\:\:\mathrm{cot}\:\alpha=\frac{\mathrm{1}+\mathrm{cos}\:\beta}{\mathrm{2}+\mathrm{sin}\:\beta} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({total}\:{area}\right)={Area}\left({right}\:{half}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{of}\:{composite}\:{area}\right) \\ $$$$\frac{\pi}{\mathrm{4}}+\mathrm{1}=\mathrm{cot}\:\alpha+\mathrm{2}\left(\mathrm{2}−\mathrm{2cot}\:\alpha\right)+\frac{\beta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\left(\mathrm{cot}\:\alpha−\mathrm{1}\right)\mathrm{sin}\:\beta \\ $$$${say}\:\:\mathrm{cot}\:\alpha={t} \\ $$$$\frac{\pi}{\mathrm{4}}+\mathrm{1}={t}+\mathrm{4}−\mathrm{4}{t}+\frac{\beta}{\mathrm{2}}−\left({t}−\mathrm{1}\right)\mathrm{sin}\:\beta \\ $$$$\Rightarrow \\ $$$$\:\frac{\pi}{\mathrm{4}}+\mathrm{1}=\mathrm{4}+\frac{\beta}{\mathrm{2}}+\mathrm{sin}\:\beta−\left(\mathrm{3}+\mathrm{sin}\:\beta\right){t} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{3}+\mathrm{sin}\:\beta\right)\left(\frac{\mathrm{1}+\mathrm{cos}\:\beta}{\mathrm{2}+\mathrm{sin}\:\beta}\right)−\frac{\beta}{\mathrm{2}}−\mathrm{sin}\:\beta \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{3}−\frac{\pi}{\mathrm{4}} \\ $$$${this}\:{eq}.\:{yields}\:\beta\:\:{and}\:{then} \\ $$$$\:\:\:\:\alpha=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}+\mathrm{sin}\:\beta}{\mathrm{1}+\mathrm{cos}\:\beta}\right)\:. \\ $$
Answered by mr W last updated on 05/Nov/19
Commented by mr W last updated on 05/Nov/19
tan γ=2 ⇒sin γ=(2/( (√5))) ⇒cos γ=(1/( (√5))) AC=(√5)R ΔACB=(((√5)R^2 sin (π−γ+β))/2)=(((√5)R^2 sin (γ−β))/2) (((√5)R^2 sin (γ−β))/2)+R^2 +(((π−β)R^2 )/2)=(1/2)(4R^2 +((πR^2 )/2)) (√5)sin (γ−β)−β=2−(π/2) 2 cos β−sin β−β=2−(π/2) ⇒β=0.6192 tan α=((2R+R sin β)/(R+R cos β))=((2+sin β)/(1+cos β)) ⇒α=54.888°
$$\mathrm{tan}\:\gamma=\mathrm{2}\:\Rightarrow\mathrm{sin}\:\gamma=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{cos}\:\gamma=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$${AC}=\sqrt{\mathrm{5}}{R} \\ $$$$\Delta{ACB}=\frac{\sqrt{\mathrm{5}}{R}^{\mathrm{2}} \mathrm{sin}\:\left(\pi−\gamma+\beta\right)}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}{R}^{\mathrm{2}} \mathrm{sin}\:\left(\gamma−\beta\right)}{\mathrm{2}} \\ $$$$\frac{\sqrt{\mathrm{5}}{R}^{\mathrm{2}} \mathrm{sin}\:\left(\gamma−\beta\right)}{\mathrm{2}}+{R}^{\mathrm{2}} +\frac{\left(\pi−\beta\right){R}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{R}^{\mathrm{2}} +\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\sqrt{\mathrm{5}}\mathrm{sin}\:\left(\gamma−\beta\right)−\beta=\mathrm{2}−\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{2}\:\mathrm{cos}\:\beta−\mathrm{sin}\:\beta−\beta=\mathrm{2}−\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\beta=\mathrm{0}.\mathrm{6192} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}{R}+{R}\:\mathrm{sin}\:\beta}{{R}+{R}\:\mathrm{cos}\:\beta}=\frac{\mathrm{2}+\mathrm{sin}\:\beta}{\mathrm{1}+\mathrm{cos}\:\beta} \\ $$$$\Rightarrow\alpha=\mathrm{54}.\mathrm{888}° \\ $$
Commented by ajfour last updated on 05/Nov/19
Thanks mrW Sir, hope my answer comes the same; haven′t checked yet..
$${Thanks}\:{mrW}\:{Sir},\:{hope}\:{my}\:{answer} \\ $$$${comes}\:{the}\:{same};\:{haven}'{t}\:{checked} \\ $$$${yet}.. \\ $$

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