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Question Number 143443 by mnjuly1970 last updated on 14/Jun/21

Commented by amin96 last updated on 14/Jun/21

$$?$$$$$$

Answered by mr W last updated on 14/Jun/21

$$sayR=radius$$$$\cos \alpha =\frac{R}{b}$$$${(b\sin \alpha +a\cos \alpha )}^2+{\left(a\sin \alpha \right)}^2=R^2$$$$b^2{\sin }^2\alpha +2ab\sin \alpha \cos \alpha +a^2=b^2{\cos }^2\alpha$$$$a^2=b(b\cos 2\alpha -a\sin 2\alpha )$$$$a^2=b\sqrt{a^2+b^2}(\cos \phi \cos 2\alpha -\sin \phi \sin 2\alpha )$$$$\frac{a^2}{b\sqrt{a^2+b^2}}=\cos (\phi +2\alpha )$$$$2\alpha ={\cos }^{-1}\frac{a^2}{b\sqrt{a^2+b^2}}-{\cos }^{-1}\frac{b^2}{b\sqrt{a^2+b^2}}$$$$\cos 2\alpha =\frac{a(ab+\sqrt{a^2{b}^2+b^4-a^4})}{b(a^2+b^2)}$$$$\Rightarrow \tan \alpha =\sqrt{\frac{b^3-a\sqrt{a^2{b}^2+b^4-a^4}}{2a^{2}b+b^3+a\sqrt{a^2{b}^2+b^4-a^4}}}$$

Commented by mnjuly1970 last updated on 14/Jun/21

$$thanksalotMrW....$$

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