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Question Number 205808 by cortano12 last updated on 31/Mar/24

Commented by A5T last updated on 02/Apr/24

$$Iguessthisisnotuniquelydefined.$$

Commented by A5T last updated on 02/Apr/24

$$Yea,{that}^{\prime }strue,makesitunique.$$

Commented by mr W last updated on 02/Apr/24

$$ADshouldtangentthesemi-circle.$$

Answered by mr W last updated on 31/Mar/24

Commented by mr W last updated on 31/Mar/24

$$\tan \frac{\alpha }{2}=\frac{R}{2x}$$$$\sin \alpha =\frac{x}{R}$$$$\frac{x}{R}=\frac{2\times \frac{R}{2x}}{1+{\left(\frac{R}{2x}\right)}^2}$$$$sayt=\frac{x}{R}$$$$t=\frac{\frac{1}{t}}{1+\frac{1}{4t^2}}=\frac{4t}{4t^2+1}$$$$\Rightarrow t=\frac{\sqrt{3}}{2}$$$$\tan \theta =\frac{2x}{2R}=\frac{x}{R}=t=\frac{\sqrt{3}}{2}$$$$\Rightarrow \sin \theta =\frac{\sqrt{3}}{\sqrt{2^2+{\left(\sqrt{3}\right)}^2}}=\sqrt{\frac{3}{7}}$$

Answered by A5T last updated on 02/Apr/24

$$AD=AC=2x;LetADandCEmeetatF$$$$Then,\frac{FE}{FC=FE+CE}=\frac{DE}{AC}=\frac{1}{2}\Rightarrow 2FE=FE+CE$$$$\Rightarrow FE=CE;\frac{AD}{AF}=\frac{1}{2}\Rightarrow AF=4x$$$$\Rightarrow FC=\sqrt{{AF}^2-{AC}^2}=\sqrt{16x^2-4x^2}=2x\sqrt{3}\Rightarrow CE=x\sqrt{3}$$$$\left(x\sqrt{3}\right)(2r-x\sqrt{3})=x^2\Rightarrow (2r=\frac{x}{\sqrt{3}}+x\sqrt{3})=\frac{4x}{\sqrt{3}}$$$$sin\theta =\frac{AC=2x}{\sqrt{\frac{16x^2}{3}+4x^2}}=\frac{2x}{x\sqrt{\frac{28}{3}}}=\frac{2\sqrt{3}}{\sqrt{28}=2\sqrt{7}}=\frac{\sqrt{3}}{\sqrt{7}}$$

Commented by A5T last updated on 02/Apr/24

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