Question Number 154353 by liberty last updated on 17/Sep/21
Commented by mr W last updated on 17/Sep/21
$${r}=\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta \\ $$$$\mathrm{2}{r}\:\mathrm{cos}\:\theta=\mathrm{3} \\ $$$$\mathrm{4}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta=\mathrm{3} \\ $$$$\mathrm{sin}\:\mathrm{2}\theta=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{2}\theta=\mathrm{60}° \\ $$$$\Rightarrow\theta=\mathrm{30}° \\ $$$$\mathrm{4}\:\mathrm{cos}\:\theta=\mathrm{2}\sqrt{\mathrm{3}}=\mathrm{2}{r}\:\checkmark \\ $$
Commented by amin96 last updated on 17/Sep/21
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Commented by amin96 last updated on 17/Sep/21
$$\mathrm{4}{r}^{\mathrm{2}} =\mathrm{16}−{a}^{\mathrm{2}} \:\:\Rightarrow\:\:\:\mathrm{2}{r}=\sqrt{\mathrm{16}−{a}^{\mathrm{2}} }\:\:\frac{{a}}{\mathrm{4}}=\frac{{a}+{r}}{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$${a}=\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\Rightarrow\mathrm{2}{r}=\sqrt{\mathrm{16}−\frac{\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{3}}}\:\:\:\:\mathrm{4}{r}^{\mathrm{2}} =\mathrm{16}−\frac{\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\frac{\mathrm{16}{r}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{16}\:\:\:\:{r}=\sqrt{\mathrm{3}}\:\:\:\:\:\:\:{a}=\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{3}}}\:\:\:\:{a}=\mathrm{2} \\ $$$$\mathrm{sin}\:\left(\theta\right)=\frac{{a}}{\mathrm{4}}=\frac{\mathrm{2}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\theta=\mathrm{30}° \\ $$
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
$${a}=\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{3}}}\:\:\:\:{r}=\sqrt{\mathrm{3}}\:\:\:\:\Rightarrow{a}=\mathrm{2}\:\:\:\:\:\:{sin}\theta=\frac{{a}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\theta=\mathrm{30}° \\ $$