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Question Number 171574 by cortano1 last updated on 18/Jun/22

Answered by mr W last updated on 18/Jun/22

Commented by mr W last updated on 18/Jun/22

$$\sin \beta =\frac{b}{2r}=\frac{b}{d}$$$$\sin \alpha =\frac{a}{2r}=\frac{a}{d}$$$$\alpha +\beta =\theta$$$$\cos (\alpha +\beta )=\cos \theta$$$$\sqrt{(1-\frac{a^2}{d^2})(1-\frac{b^2}{d^2})}-\frac{a}{d}\times \frac{b}{d}=\cos \theta$$$$(d^2-a^2)(d^2-b^2)={(ab+d^2\cos \theta )}^2$$$${\sin }^2\theta d^2=a^2+b^2+2ab\cos \theta$$$$d=\frac{\sqrt{a^2+b^2+2ab\cos \theta }}{\sin \theta }$$$$\Rightarrow r=\frac{\sqrt{a^2+b^2+2ab\cos \theta }}{2\sin \theta }$$$$=\frac{\sqrt{3^2+5^2+2\times 3\times 5\cos 60°}}{2\sin 60°}=\frac{7\sqrt{3}}{3}$$

Commented by infinityaction last updated on 18/Jun/22

$$nicesir$$

Commented by Tawa11 last updated on 18/Jun/22

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

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