Question Number 108461 by ajfour last updated on 17/Aug/20
Commented by ajfour last updated on 17/Aug/20
$${If}\:{the}\:{two}\:{shaded}\:{areas}\:{are}\:{equal}, \\ $$$${find}\:\alpha\:{in}\:{terms}\:{of}\:{radius} \\ $$$${ratio}\:\:{a}/{b}. \\ $$
Answered by mr W last updated on 17/Aug/20
$${A}_{{blue}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\pi−\mathrm{2}\alpha−\mathrm{sin}\:\mathrm{2}\alpha\right) \\ $$$${A}_{{brown}} =\frac{{b}^{\mathrm{2}} }{\mathrm{2}}\left(\pi+\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\alpha\right) \\ $$$$\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\pi−\mathrm{2}\alpha−\mathrm{sin}\:\mathrm{2}\alpha\right)=\frac{{b}^{\mathrm{2}} }{\mathrm{2}}\left(\pi+\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\alpha\right) \\ $$$$\lambda=\frac{{b}}{{a}} \\ $$$$\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\pi=\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)\left(\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\alpha\right) \\ $$$$\Rightarrow\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\alpha=\frac{\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\pi}{\mathrm{1}+\lambda^{\mathrm{2}} } \\ $$
Commented by ajfour last updated on 17/Aug/20
$${Very}\:{Nice}\:{Sir},\:{i}\:{shall}\:{follow}\:{n}\:{check} \\ $$$${again},\:{not}\:{a}\:{very}\:{good}\:{question}\:{even}.. \\ $$