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Question Number 191192 by ajfour last updated on 20/Apr/23

Commented by ajfour last updated on 20/Apr/23

$$Iflowercirclehasradiusawhile$$$$upperhasradiusb,find\theta .$$

Answered by mr W last updated on 20/Apr/23

Commented by mr W last updated on 20/Apr/23

$$OA=\frac{a}{\tan \theta }$$$$AB=\sqrt{{(a+b)}^2-{(a-b)}^2}=2\sqrt{ab}$$$$OB=\frac{a}{\tan \theta }+2\sqrt{ab}=\frac{b}{\tan \frac{\theta }{2}}$$$$lett=\tan \frac{\theta }{2}$$$$\tan \theta =\frac{2t}{1-t^2}$$$$\frac{a(1-t^2)}{2t}+2\sqrt{ab}=\frac{b}{t}$$$$t^2-4\sqrt{\frac{b}{a}}t+\frac{2b}{a}-1=0$$$$\Rightarrow t=2\sqrt{\frac{b}{a}}-\sqrt{\frac{2b}{a}+1}$$$$\Rightarrow \theta =2{\tan }^{-1}(2\sqrt{\frac{b}{a}}-\sqrt{\frac{2b}{a}+1})$$

Commented by ajfour last updated on 20/Apr/23

$$OB=\frac{a}{\tan \theta }+2\sqrt{ab}=\frac{b}{\tan \frac{\theta }{2}}$$$$\Rightarrow 1-t^2+4t\sqrt{\frac{b}{a}}=2\left(\frac{b}{a}\right)$$$$\Rightarrow t^2-4t\sqrt{\frac{b}{a}}+\frac{2b}{a}-1=0$$$$t=2\sqrt{\frac{b}{a}}\pm \sqrt{\frac{2b}{a}+1}$$$$\theta ={2\tan }^{-1}(2\sqrt{\frac{b}{a}}-\sqrt{\frac{2b}{a}+1})$$$$thankssir!$$

Commented by mr W last updated on 20/Apr/23

$$yes,youarerightsir.$$

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