Question Number 205562 by mr W last updated on 24/Mar/24
Answered by A5T last updated on 24/Mar/24
Commented by A5T last updated on 24/Mar/24
$${CE}=\sqrt{\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{1}−{cos}\theta\right)}=\sqrt{\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{1}+{cos}\theta\right)} \\ $$$$\theta=\mathrm{2}×\mathrm{30}°=\mathrm{60}°\Rightarrow\sqrt{\mathrm{2}{r}^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{2}}}=\sqrt{\mathrm{2}{x}^{\mathrm{2}} ×\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\Rightarrow{r}={x}\sqrt{\mathrm{3}}\Rightarrow{tan}\alpha=\frac{{x}}{{r}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\Rightarrow\alpha=\mathrm{30}° \\ $$
Commented by mr W last updated on 26/Mar/24
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Answered by mr W last updated on 26/Mar/24
Commented by mr W last updated on 26/Mar/24
$$\mathrm{2}\alpha=\mathrm{2}×\mathrm{30}° \\ $$$$\Rightarrow\alpha=\mathrm{30}° \\ $$