Question Number 79121 by M±th+et£s last updated on 22/Jan/20
Answered by mr W last updated on 23/Jan/20
Commented by mr W last updated on 23/Jan/20
$${R}={radius}\:{of}\:{semicircle} \\ $$$$\mathrm{2}{a}={side}\:{length}\:{of}\:{equilateral}\:{triangle} \\ $$$$\lambda=\frac{{a}}{{R}} \\ $$$$\frac{\mathrm{sin}\:\frac{\pi}{\mathrm{3}}}{{R}}=\frac{\mathrm{sin}\:\left(\pi−\beta−\frac{\pi}{\mathrm{3}}\right)}{{a}}=\frac{\mathrm{cos}\:\left(\beta−\frac{\pi}{\mathrm{6}}\right)}{{a}} \\ $$$$\mathrm{cos}\:\left(\beta−\frac{\pi}{\mathrm{6}}\right)=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}{R}}=\frac{\sqrt{\mathrm{3}}\lambda}{\mathrm{2}} \\ $$$$\Rightarrow\beta=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}\lambda}{\mathrm{2}}+\frac{\pi}{\mathrm{6}} \\ $$$$\mathrm{sin}\:\beta=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}\lambda}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{4}−\mathrm{3}\lambda^{\mathrm{2}} }}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}\left(\lambda+\sqrt{\mathrm{4}−\mathrm{3}\lambda^{\mathrm{2}} }\right)}{\mathrm{4}} \\ $$$$\alpha=\frac{\pi}{\mathrm{2}}−\beta=\frac{\pi}{\mathrm{3}}−\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}\lambda}{\mathrm{2}} \\ $$$${A}=\mathrm{2}\left(\frac{\alpha{R}^{\mathrm{2}} }{\mathrm{2}}+\frac{{Ra}\:\mathrm{sin}\:\beta}{\mathrm{2}}\right) \\ $$$${A}={R}^{\mathrm{2}} \left(\alpha+\lambda\:\mathrm{sin}\:\beta\right) \\ $$$${area}\:{of}\:{triangle}\:=\sqrt{\mathrm{3}}{a}^{\mathrm{2}} =\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} {R}^{\mathrm{2}} ={A}+{S}\:\:..\left({i}\right) \\ $$$${area}\:{of}\:{semicircle}\:=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}={A}+\mathrm{2}{S}\:\:..\left({ii}\right) \\ $$$$\mathrm{2}×\left({i}\right)−\left({ii}\right): \\ $$$$\Rightarrow\mathrm{2}\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} {R}^{\mathrm{2}} −\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}={A}={R}^{\mathrm{2}} \left(\alpha+\lambda\:\mathrm{sin}\:\beta\right) \\ $$$$\Rightarrow\mathrm{2}\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} −\frac{\pi}{\mathrm{2}}=\alpha+\lambda\:\mathrm{sin}\:\beta \\ $$$$\Rightarrow\mathrm{2}\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} −\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{3}}−\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}\lambda}{\mathrm{2}}+\frac{\lambda\sqrt{\mathrm{3}}\left(\lambda+\sqrt{\mathrm{4}−\mathrm{3}\lambda^{\mathrm{2}} }\right)}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{7}\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} −\lambda\sqrt{\mathrm{3}\left(\mathrm{4}−\mathrm{3}\lambda^{\mathrm{2}} \right)}+\mathrm{4}\:\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}\lambda}{\mathrm{2}}=\frac{\mathrm{10}\pi}{\mathrm{3}} \\ $$$$\Rightarrow\lambda\approx\mathrm{0}.\mathrm{894034} \\ $$$$ \\ $$$${from}\:\left({ii}\right): \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}−{A}\right)=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}−\frac{{A}}{\mathrm{2}} \\ $$$$\frac{{S}}{{A}}=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}×\left(\mathrm{2}\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} {R}^{\mathrm{2}} −\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}\right)}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{S}}{{A}}=\frac{\pi}{\mathrm{8}\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} −\mathrm{2}\pi}−\frac{\mathrm{1}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{155564} \\ $$
Commented by mr W last updated on 23/Jan/20
$${an}\:{exact}\:{solution}\:{is}\:{not}\:{possible}. \\ $$
Commented by M±th+et£s last updated on 23/Jan/20
$${god}\:{bless}\:{you}\:{sir}\:.\:{nice}\:{job} \\ $$
Commented by M±th+et£s last updated on 23/Jan/20
Commented by M±th+et£s last updated on 23/Jan/20
$${how}\:{can}\:{you}\:{find}\:{sin}\left({B}\right)\:{from}\:{B}\:{valuep} \\ $$