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Question Number 207731 by efronzo1 last updated on 24/May/24

$$\downharpoonleft \underset{―}{}$$

Answered by A5T last updated on 24/May/24

Commented by A5T last updated on 24/May/24

$$\mathrm{\angle }ACE=\theta \Rightarrow \mathrm{\angle }BCE=90-\theta \Rightarrow \mathrm{\angle }CEB=60+\theta$$$$\frac{sin\theta }{\frac{3r}{2}}=\frac{sin60}{14}\Rightarrow sin\theta =\frac{3\sqrt{3}r}{56}...\left(i\right)$$$$\frac{sin(90-\theta )}{\frac{r}{2}}=\frac{sin30}{14}\Rightarrow cos\theta =\frac{r}{56}...\left(ii\right)$$$$\frac{\left(i\right)}{\left(ii\right)}\Rightarrow tan\theta =3\sqrt{3}\Rightarrow \theta ={tan}^{-1}\left(3\sqrt{3}\right)$$$$\left[BCE\right]=\frac{1}{2}\times \frac{r}{2}\times 14\times sin(60+\theta )$$$$=7\times 28cos\left({tan}^{-1}\left(3\sqrt{3}\right)\right)sin(60+{tan}^{-1}\left(3\sqrt{3}\right))$$$$\approx 24.249$$

Answered by A5T last updated on 24/May/24

$$sin\theta =\frac{3\sqrt{3}r}{56};cos\theta =\frac{r}{56}\Rightarrow sin\theta =3\sqrt{3}cos\theta$$$$\Rightarrow sin\theta =3\sqrt{3(1-{sin}^2\theta )}\Rightarrow {sin}^2\theta =27(1-{sin}^2\theta )$$$$\Rightarrow 28{sin}^2\theta =27\Rightarrow sin\theta =\frac{3\sqrt{21}}{14}$$$$\Rightarrow r=\frac{56sin\theta }{3\sqrt{3}}=\frac{12\sqrt{21}}{3\sqrt{3}}=4\sqrt{7}$$$$sin(60+\theta )=sin60cos\theta +cos60sin\theta$$$$=\frac{\sqrt{3}}{2}\times \frac{\sqrt{7}}{14}+\frac{1}{2}\times \frac{3\sqrt{21}}{14}=\frac{\sqrt{21}}{7}$$$$\Rightarrow \left[BCE\right]=\frac{1}{2}\times 14\times 2\sqrt{7}\times \frac{\sqrt{21}}{7}=14\sqrt{3}$$

Commented by A5T last updated on 24/May/24

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