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Question Number 118872 by ZiYangLee last updated on 20/Oct/20

Commented by ZiYangLee last updated on 20/Oct/20

$$P\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{in}\mathrm{square}ABCD.$$$$\mathrm{Given}\mathrm{that}AP=7,PB=6,CP=11\mathrm{and}$$$$\mathrm{\angle }APB\mathrm{is}\theta .\mathrm{Find}\tan \theta .$$

Answered by 1549442205PVT last updated on 20/Oct/20

Commented by 1549442205PVT last updated on 21/Oct/20

$$\mathrm{Given}\mathrm{AE}=7,\mathrm{BE}=6,\mathrm{CE}=11.\mathrm{Denote}$$$$\widehat{\mathrm{ECF}}=\alpha ,\widehat{\mathrm{EBF}}=\beta ,\mathrm{AB}=\mathrm{BC}=\mathrm{a}.\mathrm{Then}$$$$\mathrm{CF}=11\cos \alpha ,\mathrm{BF}=6\cos \beta .\mathrm{BG}=\mathrm{EF}=6\sin \beta$$$$\mathrm{AG}=\sqrt{{\mathrm{AE}}^2-{\mathrm{EG}}^2}=\sqrt{{\mathrm{AE}}^2-{\mathrm{BF}}^2}$$$$=\sqrt{49-{36\cos }^2\beta }$$$$\mathrm{AG}+\mathrm{BG}=\mathrm{BF}+\mathrm{CF}=\mathrm{a}.\mathrm{Therefore},$$$$\sqrt{49-{36\cos }^2\beta }+6\sin \beta =6\cos \beta +11\cos \alpha (\ast )$$$$\mathrm{On}\mathrm{the}\mathrm{other}\mathrm{hand},\mathrm{EF}=11\sin \alpha =6\sin \beta$$$$\Rightarrow \sin \alpha =\frac{6\sin \beta }{11}\Rightarrow {\sin }^2\alpha =\frac{{36\sin }^2\beta }{121}$$$$\cos \alpha =\frac{\sqrt{121-{36\sin }^2\beta }}{11}.\mathrm{Replace}\mathrm{into}(\ast )$$$$\mathrm{we}\mathrm{get}$$$$\sqrt{49-{36\cos }^2\beta }+6\sin \beta =6\cos \beta +\sqrt{121-{36\sin }^2\beta }$$$$\sqrt{49-{36\cos }^2\beta }-\sqrt{121-{36\sin }^2\beta }=6\cos \beta -6\sin \beta$$$$\mathrm{Squaring}\mathrm{two}\mathrm{sides}\mathrm{we}\mathrm{get}$$$$170-36-2\sqrt{(49-{36\cos }^2\beta )(121-{36\sin }^2\beta )}$$$$=36-72\sin \beta \cos \beta .\mathrm{Note}\sin 2\beta =2\sin \beta \cos \beta$$$$49+18\sin 2\beta =\sqrt{5929-36({121\cos }^2\beta +{49\sin }^2\beta )+{324\sin }^{2}2\beta }$$$$\mathrm{Squaring}\mathrm{two}\mathrm{sides}\mathrm{again}\mathrm{we}\mathrm{get}$$$$2401+1764\sin 2\beta +{324\sin }^{2}2\beta =$$$$5929-36.49-36.{72\cos }^2\beta +{324\sin }^{2}2\beta$$$$\iff 1764-{2592\cos }^2\beta =1764\sin 2\beta \iff$$$$468-1296\cos 2\beta =1764\sin 2\beta (\mathrm{since}{2\cos }^2\beta =1+\cos 2\beta )$$$$\iff 13=36\cos 2\beta +49\sin 2\beta .\mathrm{Put}\mathrm{t}=\tan \beta$$$$\mathrm{we}\mathrm{have}36\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}+49\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}=13$$$$\iff {36\mathrm{t}}^2-98\mathrm{t}-36+{13\mathrm{t}}^2+13=0$$$$\iff {49\mathrm{t}}^2-98\mathrm{t}-23=0$$$${\mathrm{\Delta }}^{\prime }={49}^2+49.23=49.72={(7.6\sqrt{2})}^2$$$$\mathrm{t}=\tan \beta =\frac{49+42\sqrt{2}}{49}=\frac{7+6\sqrt{2}}{7}\Rightarrow \beta \approx 65°{40}^{\prime }30$$$$\cos \beta =\frac{1}{\sqrt{1+{\tan }^2\beta }}=\frac{7}{\sqrt{170+84\sqrt{2}}},$$$$,\mathrm{EG}=6\cos \beta =\frac{42}{\sqrt{170+84\sqrt{2}}}\left(1\right)$$$$\mathrm{Put}\widehat{\mathrm{BEG}}={\theta }_1,\widehat{\mathrm{AEG}}={\theta }_2.\mathrm{Then}\mathrm{we}\mathrm{have}$$$$\tan {\theta }_1=\frac{\mathrm{BG}}{\mathrm{EG}}=\frac{\sqrt{36-{\mathrm{EG}}^2}}{\mathrm{EG}}=\sqrt{\frac{36}{{\mathrm{EG}}^2}-1}$$$$\tan {\theta }_2=\frac{\mathrm{AG}}{\mathrm{EG}}=\frac{\sqrt{49-{\mathrm{EG}}^2}}{\mathrm{EG}}=\sqrt{\frac{49}{{\mathrm{EG}}^2}-1}$$$$\mathrm{From}\mathrm{that}\mathrm{and}\left(1\right)\mathrm{we}\mathrm{get}$$$$\tan \theta =\tan ({\theta }_1+{\theta }_2)=\frac{\tan {\theta }_1+\tan {\theta }_2}{1-\tan {\theta }_1\tan {\theta }_2}$$$$=\frac{\sqrt{\frac{36}{{\mathrm{x}}^2}-1}+\sqrt{\frac{49}{{\mathrm{x}}^2}-1}}{1-\sqrt{(\frac{36}{{\mathrm{x}}^2}-1)(\sqrt{\frac{49}{{\mathrm{x}}^2}-1)}}}$$$$\tan \theta =\frac{\mathrm{x}(\sqrt{36-{\mathrm{x}}^2}+\sqrt{49-{\mathrm{x}}^2})}{{\mathrm{x}}^2-\sqrt{(36-{\mathrm{x}}^2)(49-{\mathrm{x}}^2)}},\mathrm{x}=\mathrm{EG}$$$$\mathbf{We}\mathbf{find}\mathbf{out}\mathbf{tan}\mathit{\theta }=-1\Rightarrow \mathit{\theta }=135°$$

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