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Question Number 78289 by ajfour last updated on 15/Jan/20

Commented by ajfour last updated on 15/Jan/20

$$Find\theta ,given\alpha .$$

Answered by mr W last updated on 15/Jan/20

$$R=radius$$$$\frac{2R\cos \alpha }{\cos \theta \sin (2\theta +\alpha )}=\frac{R}{\sin \theta }$$$$\frac{\sin (2\theta +\alpha )}{\cos \alpha }=\frac{2\sin \theta }{\cos \theta }$$$$\frac{\sin \left(2\theta \right)\cos \alpha +\cos \left(2\theta \right)\sin \alpha }{\cos \alpha }=2\tan \theta$$$$\sin \left(2\theta \right)+\tan \alpha \cos \left(2\theta \right)=2\tan \theta$$$$\tan \alpha (2{\cos }^2\theta -1)=2\tan \theta (1-{\cos }^2\theta )$$$$\tan \alpha (\frac{2}{1+{\tan }^2\theta }-1)=2\tan \theta (1-\frac{1}{1+{\tan }^2\theta })$$$$\tan \alpha \left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)=2\tan \theta \left(\frac{{\tan }^2\theta }{1+{\tan }^2\theta }\right)$$$$\tan \alpha (1-{\tan }^2\theta )=2{\tan }^3\theta$$$$\Rightarrow \frac{1}{{\tan }^3\theta }-\frac{1}{\tan \theta }-\frac{2}{\tan \alpha }=0$$$$lett=\frac{1}{\tan \theta },a=\frac{1}{\tan \alpha }$$$$\Rightarrow t^3-t-2a=0$$$$\Rightarrow t=\sqrt[3]{\sqrt{a^2-\frac{1}{27}}+a}-\sqrt[3]{\sqrt{a^2-\frac{1}{27}}-a}$$$$\Rightarrow \frac{1}{\tan \theta }=\sqrt[3]{\sqrt{\frac{1}{{\tan }^2\alpha }-\frac{1}{27}}+\frac{1}{\tan \alpha }}-\sqrt[3]{\sqrt{\frac{1}{{\tan }^2\alpha }-\frac{1}{27}}-\frac{1}{\tan \alpha }}$$$$\Rightarrow \theta =\frac{\pi }{2}-{\tan }^{-1}(\sqrt[3]{\sqrt{\frac{1}{{\tan }^2\alpha }-\frac{1}{27}}+\frac{1}{\tan \alpha }}-\sqrt[3]{\sqrt{\frac{1}{{\tan }^2\alpha }-\frac{1}{27}}-\frac{1}{\tan \alpha }})$$

Commented by ajfour last updated on 16/Jan/20

$$ThankyouSir,butsoon$$$$Iamgoingtofindawayout$$$$ofthiscumbersomeanswer.$$

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