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Question Number 172437 by infinityaction last updated on 26/Jun/22

Answered by mr W last updated on 27/Jun/22

$$R=radiusofsemicircle$$$$r=radiusofcircle$$$$O=centerofsemicircle$$$$AC=2R\cos \theta$$$$AN=AM=\frac{r}{\tan \frac{\theta }{2}}$$$$ON=\frac{r}{\tan \frac{\theta }{2}}-R$$$${(\frac{r}{\tan \frac{\theta }{2}}-R)}^2={(R-r)}^2-r^2$$$$\frac{r}{{\tan }^2\frac{\theta }{2}}=2R(\frac{1}{\tan \frac{\theta }{2}}-1)$$$$\Rightarrow r=2R(1-\tan \frac{\theta }{2})\tan \frac{\theta }{2}$$$$a=2R\cos \theta -\frac{r}{\tan \frac{\theta }{2}}=2R(\cos \theta -1+\tan \frac{\theta }{2})$$$$b=2R-\frac{r}{\tan \frac{\theta }{2}}=2R\tan \frac{\theta }{2}$$$$\frac{a}{b}=\frac{\cos \theta -1+\tan \frac{\theta }{2}}{\tan \frac{\theta }{2}}$$$$=1-\frac{1-\cos \theta }{\tan \frac{\theta }{2}}$$$$=1-\frac{2{\sin }^2\frac{\theta }{2}}{\tan \frac{\theta }{2}}$$$$=1-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$$$$=1-\sin \theta ✓$$

Commented by infinityaction last updated on 27/Jun/22

$$thankyousir$$

Commented by Tawa11 last updated on 27/Jun/22

$$\mathrm{Great}\mathrm{sir}$$

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