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Question Number 40031 by ajfour last updated on 15/Jul/18

Commented by ajfour last updated on 16/Jul/18

$$CEandDF\mathrm{\perp }AF.Find\theta .$$

Answered by ajfour last updated on 21/Jul/18

$$letAbeoriginandx-axisleftwards.$$$$AF=\sin \theta$$$$x_C=\sin \theta -\cos \theta$$$$eq.ofBF:y=-\frac{\sin \theta }{\sin \theta +\cos \theta }(x-\sin \theta )$$$$eq.ofAD:y=\frac{\cos \theta }{\sin \theta }x$$$$sofor{x}_E:$$$$-\frac{\sin \theta }{\sin \theta +\cos \theta }(x_E-\sin \theta )=\frac{\cos \theta }{\sin \theta }{x}_E$$$$\Rightarrow x_E(\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\sin \theta +\cos \theta })=\frac{\sin {}^2\theta }{\sin \theta +\cos \theta }$$$$\Rightarrow x_E=\frac{\sin {}^3\theta }{\sin \theta \cos \theta +1}$$$$as{x}_E=x_{C}wehave$$$$\sin \theta -\cos \theta =\frac{\sin {}^3\theta }{\sin \theta \cos \theta +1}$$$$\Rightarrow \sin {}^2\theta \cos \theta +\sin \theta -\sin \theta \cos {}^2\theta$$$$-\cos \theta =\sin {}^3\theta$$$$\Rightarrow \sin {}^2\theta \cos \theta +\sin \theta -\sin \theta \cos {}^2\theta$$$$-\cos \theta =\sin \theta -\sin \theta \cos {}^2\theta$$$$\Rightarrow \sin {}^2\theta \cos \theta =\cos \theta$$$$\Rightarrow \sin \theta =1or\theta =\frac{\pi }{2}.$$

Commented by ajfour last updated on 21/Jul/18

$$butihadobtainedsomeacute$$$$anglevaluefor\theta when{i}^{\prime }dsolved$$$$before;dunnowwhyi{couldn}^{\prime }t$$$$obtainthesamenow..$$

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