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Question Number 68289 by Rasheed.Sindhi last updated on 08/Sep/19

A circle is divided into two equal parts By An arc with center on the circle. Determine (a) The length of the arc (b)The ratio in which the arc divides the diameter meeting the center of the arc.

$$\mathrm{A}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{into}\:\mathrm{two}\:\mathrm{equal}\:\mathrm{parts} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{By} \\ $$$$\:\mathrm{An}\:\mathrm{arc}\:\mathrm{with}\:\mathrm{center}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circle}. \\ $$$$\mathcal{D}{etermine} \\ $$$$\:\:\left({a}\right)\:{The}\:{length}\:{of}\:{the}\:{arc} \\ $$$$\:\:\left({b}\right){The}\:{ratio}\:{in}\:{which}\:{the}\:{arc} \\ $$$$\:\:\:\:\:\:\:\:{divides}\:{the}\:{diameter}\: \\ $$$$\:\:\:\:\:\:\:\:{meeting}\:{the}\:{center}\:{of}\:{the}\:{arc}. \\ $$

Answered by mr W last updated on 08/Sep/19

Commented by mr W last updated on 09/Sep/19

let λ=(r/R) cos β=(r/(2R))=(λ/2) cos 2β=2((λ/2))^2 −1=((λ^2 −2)/2) sin 2β=((√(2^2 −(λ^2 −2)^2 ))/2)=((λ(√(4−λ^2 )))/2) α+2β=π 2α=2π−4β sin 2α=−sin 4β=−2 sin 2β cos 2β =−2×((λ(√(4−λ^2 )))/2)×((λ^2 −2)/2) =((λ(2−λ^2 )(√(4−λ^2 )))/2) (r^2 /2)(2β−sin 2β)+(R^2 /2)(2α−sin 2α)=((πR^2 )/2) ⇒λ^2 (2 cos^(−1) (λ/2)−((λ(√(4−λ^2 )))/2))+2π−4 cos^(−1) (λ/2)−((λ(2−λ^2 )(√((2−λ)(2+λ))))/2)=π ⇒2(2−λ^2 )cos^(−1) (λ/2)+λ(√(4−λ^2 ))=π ⇒λ≈1.1587 the diameter is divided in ratio 0.8413 : 1.1587

$${let}\:\lambda=\frac{{r}}{{R}} \\ $$$$\mathrm{cos}\:\beta=\frac{{r}}{\mathrm{2}{R}}=\frac{\lambda}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\mathrm{2}\beta=\mathrm{2}\left(\frac{\lambda}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1}=\frac{\lambda^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\mathrm{2}\beta=\frac{\sqrt{\mathrm{2}^{\mathrm{2}} −\left(\lambda^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} }}{\mathrm{2}}=\frac{\lambda\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\alpha+\mathrm{2}\beta=\pi \\ $$$$\mathrm{2}\alpha=\mathrm{2}\pi−\mathrm{4}\beta \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha=−\mathrm{sin}\:\mathrm{4}\beta=−\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\beta\:\mathrm{cos}\:\mathrm{2}\beta \\ $$$$=−\mathrm{2}×\frac{\lambda\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }}{\mathrm{2}}×\frac{\lambda^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}} \\ $$$$=\frac{\lambda\left(\mathrm{2}−\lambda^{\mathrm{2}} \right)\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{2}\beta−\mathrm{sin}\:\mathrm{2}\beta\right)+\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{2}\alpha−\mathrm{sin}\:\mathrm{2}\alpha\right)=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\lambda^{\mathrm{2}} \left(\mathrm{2}\:\mathrm{cos}^{−\mathrm{1}} \frac{\lambda}{\mathrm{2}}−\frac{\lambda\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }}{\mathrm{2}}\right)+\mathrm{2}\pi−\mathrm{4}\:\mathrm{cos}^{−\mathrm{1}} \frac{\lambda}{\mathrm{2}}−\frac{\lambda\left(\mathrm{2}−\lambda^{\mathrm{2}} \right)\sqrt{\left(\mathrm{2}−\lambda\right)\left(\mathrm{2}+\lambda\right)}}{\mathrm{2}}=\pi \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{2}−\lambda^{\mathrm{2}} \right)\mathrm{cos}^{−\mathrm{1}} \frac{\lambda}{\mathrm{2}}+\lambda\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }=\pi \\ $$$$\Rightarrow\lambda\approx\mathrm{1}.\mathrm{1587} \\ $$$$ \\ $$$${the}\:{diameter}\:{is}\:{divided}\:{in}\:{ratio} \\ $$$$\mathrm{0}.\mathrm{8413}\::\:\mathrm{1}.\mathrm{1587} \\ $$

Commented by Rasheed.Sindhi last updated on 08/Sep/19

Thαnχ for this Nice Approach Sir! Could λ be obtained in terms of π.

$$\mathbb{T}\mathrm{h}\alpha\mathrm{n}\chi\:\mathrm{for}\:\mathrm{this}\:\mathbb{N}\mathrm{ice}\:\mathbb{A}\mathrm{pproach}\:\mathbb{S}\mathrm{ir}! \\ $$$$\mathrm{Could}\:\lambda\:\mathrm{be}\:\mathrm{obtained}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\pi. \\ $$

Commented by mr W last updated on 08/Sep/19

i don′t think that the equation for λ can be exactly solved.

$${i}\:{don}'{t}\:{think}\:{that}\:{the}\:{equation}\:{for}\:\lambda \\ $$$${can}\:{be}\:{exactly}\:{solved}. \\ $$

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