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Question-199723




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

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