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Question Number 40023 by ajfour last updated on 15/Jul/18

Answered by MrW3 last updated on 15/Jul/18

$$AB=2R\cos \theta$$$$AD=\frac{R}{\cos \theta }$$$$BD=AB-AD=R(2\cos \theta -\frac{1}{\cos \theta })=\frac{R\cos 2\theta }{\cos \theta }$$$$LetE=midpointofBC.$$$$BE=AB\sin \theta =R\sin 2\theta$$$$BC=2R\sin 2\theta$$$$\frac{AB}{BE}=\frac{BC}{BD}$$$$\Rightarrow \frac{2R\cos \theta }{R\sin 2\theta }=\frac{2R\sin 2\theta \cos \theta }{R\cos 2\theta }$$$$\Rightarrow \frac{1}{\sin 2\theta }=\frac{\sin 2\theta }{\cos 2\theta }$$$$\cos 2\theta ={\sin }^{2}2\theta =1-{\cos }^{2}2\theta$$$${\cos }^{2}2\theta +\cos 2\theta -1=0$$$$4{\cos }^4\theta -2{\cos }^2\theta -1=0$$$$\Rightarrow {\cos }^2\theta =\frac{1+\sqrt{5}}{4}$$$$\Rightarrow \cos \theta =\frac{\sqrt{1+\sqrt{5}}}{2}$$$$\Rightarrow \theta ={\cos }^{-1}\frac{\sqrt{1+\sqrt{5}}}{2}\approx 25.9°$$

Commented by ajfour last updated on 15/Jul/18

$$ThankyouSir,verycorrect!$$

Answered by ajfour last updated on 15/Jul/18

$$AB=2\cos \theta$$$$BC=4\cos \theta \sin \theta$$$$BD=4\cos \theta \sin {}^2\theta$$$$BD+AD=AB$$$$\Rightarrow 4\cos \theta \sin {}^2\theta +\frac{1}{\cos \theta }=2\cos \theta$$$$let\cos \theta =t$$$$\Rightarrow 4t(1-t^2)+\frac{1}{t}=2t$$$$or4t^2(1-t^2)+1-2t^2=0$$$$\Rightarrow 4t^4-2t^2-1=0$$$$or{t}^2=\frac{1+\sqrt{5}}{4}=\cos {}^2\theta$$$$\cos \theta =\frac{\sqrt{1+\sqrt{5}}}{2}.$$

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