Question Number 199723 by cortano12 last updated on 08/Nov/23
Answered by deleteduser1 last updated on 08/Nov/23
$${Suppose}\:{ABCD}\:{is}\:{a}\:{square} \\ $$$${Through}\:{P},{let}\:{the}\:{line}\:{parallel}\:{to}\:{BC}\:{meet}\:{AB} \\ $$$${at}\:{F};{then}\:{PF}=\mathrm{8}\Rightarrow\frac{{sin}\left(\mathrm{2}\alpha\right)}{\mathrm{1}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\Rightarrow{cos}\left(\mathrm{2}\alpha\right)=\frac{\mathrm{3}}{\mathrm{5}}=\frac{{AF}}{\mathrm{10}}\Rightarrow{AF}=\mathrm{6}\Rightarrow{PC}=\mathrm{2} \\ $$