Question Number 84288 by jagoll last updated on 11/Mar/20
Commented by jagoll last updated on 11/Mar/20
$$\mathrm{dear}\:\mathrm{Mr}\:\mathrm{W}.\:\mathrm{please}\:\mathrm{help}\:\mathrm{me} \\ $$
Commented by jagoll last updated on 11/Mar/20
$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{the}\:\mathrm{question}\:\mathrm{not}\:\mathrm{complet} \\ $$
Commented by mr W last updated on 11/Mar/20
$${too}\:{less}\:{information}! \\ $$$${the}\:{shaded}\:{area}\:{depents}\:{on}\:{the}\:{radius} \\ $$$${of}\:{circle}\:{P}_{\mathrm{1}} .\: \\ $$
Commented by jagoll last updated on 11/Mar/20
$$\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{P}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as} \\ $$$$\mathrm{diameter}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{P}_{\mathrm{3}} \:\mathrm{sir} \\ $$
Commented by mr W last updated on 11/Mar/20
$${but}\:{i}\:{asked}\:{for}\:{radius}\:{of}\:{circle}\:{P}_{\mathrm{1}} . \\ $$
Commented by jagoll last updated on 11/Mar/20
$$\mathrm{the}\:\mathrm{original}\:\mathrm{question}\:\mathrm{circle}\:\mathrm{P}_{\mathrm{1}} \:\mathrm{becomes} \\ $$$$\mathrm{P}_{\mathrm{2}} .\:\mathrm{i}\:\mathrm{mispresent} \\ $$$$ \\ $$
Commented by mr W last updated on 11/Mar/20
$${we}\:{must}\:{know}\:{the}\:{radius}\:{both}\:{from} \\ $$$${P}_{\mathrm{1}} \:{and}\:{from}\:{P}_{\mathrm{2}} . \\ $$
Commented by jagoll last updated on 11/Mar/20
$$\mathrm{sir}\:\mathrm{w}.\:\mathrm{let}\:\mathrm{radius}\:\mathrm{circle}\:\mathrm{P}_{\mathrm{1}} \:\mathrm{is}\:\sqrt{\mathrm{5}} \\ $$$$\mathrm{what}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}\:\mathrm{sir}? \\ $$
Answered by mr W last updated on 11/Mar/20
Commented by mr W last updated on 11/Mar/20
$${BC}=\mathrm{2} \\ $$$${AB}={AC}=\sqrt{\mathrm{5}} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}\beta=\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\alpha=\pi−\mathrm{2}\beta \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha=−\mathrm{sin}\:\mathrm{4}\beta=\mathrm{2}×\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{24}}{\mathrm{25}} \\ $$$${A}_{{shade}} =\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{2}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}−\frac{\mathrm{4}}{\mathrm{5}}\right)+\frac{\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{2}\pi−\mathrm{4}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}−\frac{\mathrm{24}}{\mathrm{25}}\right)−\pi×\mathrm{1}^{\mathrm{2}} \\ $$$$=\mathrm{4}\pi−\mathrm{4}−\mathrm{6}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$=\mathrm{1}.\mathrm{923} \\ $$
Commented by jagoll last updated on 11/Mar/20
$$\mathrm{cos}\:\beta\:=\:\frac{\mathrm{5}+\mathrm{4}−\mathrm{5}}{\mathrm{2}.\sqrt{\mathrm{5}}.\mathrm{2}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\mathrm{sir}\: \\ $$$$\mathrm{by}\:\mathrm{cosine}\:\mathrm{of}\:\mathrm{law}\:? \\ $$
Commented by john santu last updated on 12/Mar/20
$$\mathrm{yes} \\ $$
Commented by mr W last updated on 12/Mar/20
$${AB}={AC} \\ $$$$\Rightarrow\angle{B}=\angle{C}=\beta \\ $$$$\mathrm{cos}\:\beta=\frac{\frac{{BC}}{\mathrm{2}}}{{AB}}=\frac{\frac{\mathrm{2}}{\mathrm{2}}}{\:\sqrt{\mathrm{5}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\:\:\left(=\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\right) \\ $$