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Question Number 123843 by mnjuly1970 last updated on 28/Nov/20

Commented by mnjuly1970 last updated on 28/Nov/20

please solve⇑⇑

$${please}\:{solve}\Uparrow\Uparrow \\ $$

Answered by mr W last updated on 28/Nov/20

Commented by mr W last updated on 28/Nov/20

say radius=R area of segment=((semicircle)/3)=((circle)/6) (R^2 /2)(π−2β−sin 2β)=((πR^2 )/6) 2β+sin 2β=((2π)/3) ⇒β=0.5863(=33.6°) PQ=2R cos β (1/2)×2R cos β×PS×sin β=((πR^2 )/6) ⇒PS=((πR)/(3 sin 2β)) ((sin (α−β))/(sin α))=((PS)/(PQ))=((πR)/(3 sin 2β×2R cos β)) cos β−((sin β)/(tan α))=(π/(6 sin 2β cos β)) tan α=((sin β)/(cos β−(π/(6 sin 2β cos β)))) tan α=((sin 2β)/(1+cos 2β−(π/(3 sin 2β)))) α=tan^(−1) (((sin 2β)/(1+cos 2β−(π/(3 sin 2β)))))=74.7°

$${say}\:{radius}={R} \\ $$$${area}\:{of}\:{segment}=\frac{{semicircle}}{\mathrm{3}}=\frac{{circle}}{\mathrm{6}} \\ $$$$\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left(\pi−\mathrm{2}\beta−\mathrm{sin}\:\mathrm{2}\beta\right)=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\mathrm{2}\beta+\mathrm{sin}\:\mathrm{2}\beta=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\Rightarrow\beta=\mathrm{0}.\mathrm{5863}\left(=\mathrm{33}.\mathrm{6}°\right) \\ $$$$ \\ $$$${PQ}=\mathrm{2}{R}\:\mathrm{cos}\:\beta \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{R}\:\mathrm{cos}\:\beta×{PS}×\mathrm{sin}\:\beta=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Rightarrow{PS}=\frac{\pi{R}}{\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\beta} \\ $$$$\frac{\mathrm{sin}\:\left(\alpha−\beta\right)}{\mathrm{sin}\:\alpha}=\frac{{PS}}{{PQ}}=\frac{\pi{R}}{\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\beta×\mathrm{2}{R}\:\mathrm{cos}\:\beta} \\ $$$$\mathrm{cos}\:\beta−\frac{\mathrm{sin}\:\beta}{\mathrm{tan}\:\alpha}=\frac{\pi}{\mathrm{6}\:\mathrm{sin}\:\mathrm{2}\beta\:\:\mathrm{cos}\:\beta} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{sin}\:\beta}{\mathrm{cos}\:\beta−\frac{\pi}{\mathrm{6}\:\mathrm{sin}\:\mathrm{2}\beta\:\:\mathrm{cos}\:\beta}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\beta−\frac{\pi}{\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\beta}} \\ $$$$\alpha=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\beta−\frac{\pi}{\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\beta}}\right)=\mathrm{74}.\mathrm{7}° \\ $$

Commented by mnjuly1970 last updated on 29/Nov/20

thank you so much master W....

$${thank}\:{you}\:{so}\:{much} \\ $$$${master}\:\:{W}.... \\ $$

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