Question Number 100644 by I want to learn more last updated on 27/Jun/20
Commented by ghiniboss last updated on 27/Jun/20
$$\mathrm{looking}\:\mathrm{for}\:\mathrm{x}? \\ $$
Answered by mr W last updated on 28/Jun/20
$${let}\:\angle{ABP}=\alpha,\:\angle{CBP}=\beta \\ $$$$\alpha+\beta=\mathrm{90}° \\ $$$$\mathrm{15}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\mathrm{2}{x}×\mathrm{9}\:\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\frac{{x}^{\mathrm{2}} −\mathrm{144}}{\mathrm{18}{x}} \\ $$$$\mathrm{12}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\mathrm{2}{x}×\mathrm{9}\:\mathrm{cos}\:\beta \\ $$$$\Rightarrow\mathrm{cos}\:\beta=\frac{{x}^{\mathrm{2}} −\mathrm{63}}{\mathrm{18}{x}}=\mathrm{sin}\:\alpha \\ $$$$\left(\frac{{x}^{\mathrm{2}} −\mathrm{144}}{\mathrm{18}{x}}\right)^{\mathrm{2}} +\left(\frac{{x}^{\mathrm{2}} −\mathrm{63}}{\mathrm{18}{x}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{4}} −\mathrm{738}{x}^{\mathrm{2}} +\mathrm{24705}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{738}\pm\mathrm{54}\sqrt{\mathrm{119}}}{\mathrm{4}} \\ $$$${x}=\sqrt{\frac{\mathrm{738}+\mathrm{54}\sqrt{\mathrm{119}}}{\mathrm{4}}}=\mathrm{18}.\mathrm{214} \\ $$$$ \\ $$$${there}\:{is}\:{an}\:{other}\:{solution},\:{if}\:{it}\:{is} \\ $$$${allowed}\:{that}\:{P}\:{can}\:{lie}\:{outside}\:{the} \\ $$$${triangle}: \\ $$$${x}=\sqrt{\frac{\mathrm{738}−\mathrm{54}\sqrt{\mathrm{119}}}{\mathrm{4}}}=\mathrm{6}.\mathrm{101} \\ $$
Commented by I want to learn more last updated on 28/Jun/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by 1549442205 last updated on 28/Jun/20
Commented by 1549442205 last updated on 28/Jun/20