Question Number 207731 by efronzo1 last updated on 24/May/24
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Answered by A5T last updated on 24/May/24
Commented by A5T last updated on 24/May/24
$$\angle{ACE}=\theta\Rightarrow\angle{BCE}=\mathrm{90}−\theta\Rightarrow\angle{CEB}=\mathrm{60}+\theta \\ $$$$\frac{{sin}\theta}{\frac{\mathrm{3}{r}}{\mathrm{2}}}=\frac{{sin}\mathrm{60}}{\mathrm{14}}\Rightarrow{sin}\theta=\frac{\mathrm{3}\sqrt{\mathrm{3}}{r}}{\mathrm{56}}…\left({i}\right) \\ $$$$\frac{{sin}\left(\mathrm{90}−\theta\right)}{\frac{{r}}{\mathrm{2}}}=\frac{{sin}\mathrm{30}}{\mathrm{14}}\Rightarrow{cos}\theta=\frac{{r}}{\mathrm{56}}…\left({ii}\right) \\ $$$$\frac{\left({i}\right)}{\left({ii}\right)}\Rightarrow{tan}\theta=\mathrm{3}\sqrt{\mathrm{3}}\Rightarrow\theta={tan}^{−\mathrm{1}} \left(\mathrm{3}\sqrt{\mathrm{3}}\right) \\ $$$$\left[{BCE}\right]=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{r}}{\mathrm{2}}×\mathrm{14}×{sin}\left(\mathrm{60}+\theta\right) \\ $$$$=\mathrm{7}×\mathrm{28}{cos}\left({tan}^{−\mathrm{1}} \left(\mathrm{3}\sqrt{\mathrm{3}}\right)\right){sin}\left(\mathrm{60}+{tan}^{−\mathrm{1}} \left(\mathrm{3}\sqrt{\mathrm{3}}\right)\right) \\ $$$$\approx\mathrm{24}.\mathrm{249} \\ $$
Answered by A5T last updated on 24/May/24
$${sin}\theta=\frac{\mathrm{3}\sqrt{\mathrm{3}}{r}}{\mathrm{56}};{cos}\theta=\frac{{r}}{\mathrm{56}}\Rightarrow{sin}\theta=\mathrm{3}\sqrt{\mathrm{3}}{cos}\theta \\ $$$$\Rightarrow{sin}\theta=\mathrm{3}\sqrt{\mathrm{3}\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right)}\Rightarrow{sin}^{\mathrm{2}} \theta=\mathrm{27}\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right) \\ $$$$\Rightarrow\mathrm{28}{sin}^{\mathrm{2}} \theta=\mathrm{27}\Rightarrow{sin}\theta=\frac{\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{56}{sin}\theta}{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{12}\sqrt{\mathrm{21}}}{\mathrm{3}\sqrt{\mathrm{3}}}=\mathrm{4}\sqrt{\mathrm{7}} \\ $$$${sin}\left(\mathrm{60}+\theta\right)={sin}\mathrm{60}{cos}\theta+{cos}\mathrm{60}{sin}\theta \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}}=\frac{\sqrt{\mathrm{21}}}{\mathrm{7}} \\ $$$$\Rightarrow\left[{BCE}\right]=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{14}×\mathrm{2}\sqrt{\mathrm{7}}×\frac{\sqrt{\mathrm{21}}}{\mathrm{7}}=\mathrm{14}\sqrt{\mathrm{3}} \\ $$
Commented by A5T last updated on 24/May/24