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Question Number 167008 by mnjuly1970 last updated on 04/Mar/22

Answered by mr W last updated on 06/Mar/22

Commented by Tawa11 last updated on 06/Mar/22

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

Commented by mr W last updated on 06/Mar/22

$$\cos \alpha =\frac{30}{2R}=\frac{15}{R}$$$$\beta =\pi -2\alpha$$$$\sin \beta =\sin 2\alpha =\frac{48}{2R}=\frac{24}{R}$$$$\cos 2\alpha =\sqrt{1-{\left(\frac{24}{R}\right)}^2}=2{\left(\frac{15}{R}\right)}^2-1$$$$1-\frac{{24}^2}{R^2}=4\times \frac{{15}^4}{R^4}-4\times \frac{{15}^2}{R^2}+1$$$$R^2=\frac{{15}^4}{{15}^2-{12}^2}$$$$\Rightarrow R=25$$$${(R-x)}^2+{48}^2={\left(2R\right)}^2$$$$\Rightarrow x=R-\sqrt{{\left(2R\right)}^2-{48}^2}=25-14=11$$

Commented by mr W last updated on 06/Mar/22

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