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Question Number 154353 by liberty last updated on 17/Sep/21

Commented by mr W last updated on 17/Sep/21

$$r=2\sqrt{3}\sin \theta$$$$2r\cos \theta =3$$$$4\sqrt{3}\sin \theta \cos \theta =3$$$$\sin 2\theta =\frac{\sqrt{3}}{2}$$$$2\theta =60°$$$$\Rightarrow \theta =30°$$$$4\cos \theta =2\sqrt{3}=2r✓$$

Commented by amin96 last updated on 17/Sep/21

Commented by amin96 last updated on 17/Sep/21

$$4r^2=16-a^2\Rightarrow 2r=\sqrt{16-a^2}\frac{a}{4}=\frac{a+r}{4+2\sqrt{3}}$$$$a=\frac{2r}{\sqrt{3}}\Rightarrow 2r=\sqrt{16-\frac{4r^2}{3}}4r^2=16-\frac{4r^2}{3}$$$$\frac{16r^2}{3}=16r=\sqrt{3}a=\frac{2r}{\sqrt{3}}a=2$$$$\sin \left(\theta \right)=\frac{a}{4}=\frac{2}{4}=\frac{1}{2}\theta =30°$$

Commented by Tawa11 last updated on 21/Sep/21

$$\mathrm{nice}\mathrm{sir}$$

Answered by EDWIN88 last updated on 17/Sep/21

Commented by amin96 last updated on 17/Sep/21

$$a=\frac{2r}{\sqrt{3}}r=\sqrt{3}\Rightarrow a=2sin\theta =\frac{a}{4}=\frac{1}{2}\theta =30°$$

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