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Question Number 199149 by ajfour last updated on 28/Oct/23

Commented by ajfour last updated on 28/Oct/23

$$No;rathera=1<b.FindR=f\left(b\right).$$

Answered by ajfour last updated on 28/Oct/23

Answered by mr W last updated on 29/Oct/23

Commented by mr W last updated on 29/Oct/23

$$\frac{R}{\cos \theta }-\frac{a}{\tan (\frac{\pi }{4}-\frac{\theta }{2})}=\sqrt{{(R+a)}^2-a^2}$$$$let\alpha =\frac{a}{R},\beta =\frac{b}{R}$$$$\frac{1}{\cos \theta }-\frac{\alpha }{\tan (\frac{\pi }{4}-\frac{\theta }{2})}=\sqrt{1+2\alpha }$$$$(2\alpha +1)+2\tan (\frac{\pi }{4}-\frac{\theta }{2})\sqrt{2\alpha +1}-(1+\frac{2\tan (\frac{\pi }{4}-\frac{\theta }{2})}{\cos \theta })=0$$$$\sqrt{2\alpha +1}=\sqrt{1+{\tan }^2(\frac{\pi }{4}-\frac{\theta }{2})+\frac{2\tan (\frac{\pi }{4}-\frac{\theta }{2})}{\cos \theta }}-\tan (\frac{\pi }{4}-\frac{\theta }{2})$$$$\Rightarrow \alpha \left(\theta \right)=\frac{1}{2}{[\sqrt{1+{\tan }^2(\frac{\pi }{4}-\frac{\theta }{2})+\frac{2\tan (\frac{\pi }{4}-\frac{\theta }{2})}{\cos \theta }}-\tan (\frac{\pi }{4}-\frac{\theta }{2})]}^2-\frac{1}{2}$$$$similarly$$$$\Rightarrow \beta \left(\theta \right)=\frac{1}{2}{[\sqrt{1+{\tan }^2\frac{\theta }{2}+\frac{2\tan \frac{\theta }{2}}{\sin \theta }}-\tan \frac{\theta }{2}]}^2-\frac{1}{2}$$$$from\frac{\beta \left(\theta \right)}{a\left(\theta \right)}=\frac{b}{a}wesolvefor\theta and$$$$thengetR=\frac{b}{\beta \left(\theta \right)}.$$$$$$$$example:$$$$a=1,b=3$$$$\Rightarrow \theta \approx 0.5658\Rightarrow R\approx 15.1054$$

Commented by mr W last updated on 29/Oct/23

Commented by ajfour last updated on 29/Oct/23

$$Thankyousir.Iamtryingstillfor$$$$anexactone.$$

Answered by mr W last updated on 29/Oct/23

Commented by mr W last updated on 29/Oct/23

$$equ.oftangentline:$$$$x\cos \theta +y\sin \theta -R=0$$$$centerofcircleA(\sqrt{R(R+2a)},a)$$$$centerofcircleB(b,\sqrt{R(R+2b})$$$$\sqrt{R(R+2a)}\cos \theta +a\sin \theta -R=-a$$$$b\cos \theta +\sqrt{R(R+2a)}\sin \theta -R=-b$$$$\Rightarrow \sin \theta =\frac{(R-b)\sqrt{R(R+2a)}-(R-a)b}{R\sqrt{(R+2a)(R+2b)}-ab}$$$$\Rightarrow \cos \theta =\frac{(R-a)\sqrt{R(R+2b)}-(R-b)a}{R\sqrt{(R+2a)(R+2b)}-ab}$$$${[(R-b)\sqrt{R(R+2a)}-(R-a)b]}^2+{[(R-a)\sqrt{R(R+2b)}-(R-b)a]}^2={[R\sqrt{(R+2a)(R+2b)}-ab]}^2$$$$$$$$example:$$$$\frac{b}{a}=3\Rightarrow \frac{R}{a}\approx 15.104081062050845$$

Commented by ajfour last updated on 29/Oct/23

$$Thisiswhatihadhoped!thanksSir.$$

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