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Question Number 108332 by I want to learn more last updated on 16/Aug/20

Answered by mr W last updated on 16/Aug/20

Commented by mr W last updated on 16/Aug/20

$$\cos \alpha =\frac{L^2+7^2-{11}^2}{2\times 7\times L}$$$$\sin \alpha =\cos \beta =\frac{L^2+7^2-3^2}{2\times 7\times L}$$$${\left(\frac{L^2+7^2-{11}^2}{2\times 7\times L}\right)}^2+{\left(\frac{L^2+7^2-3^2}{2\times 7\times L}\right)}^2=1$$$$L^4-({11}^2+3^2)L^2+\frac{{(7^2-{11}^2)}^2+{(7^2-3^3)}^2}{2}=0$$$$\Rightarrow L^2=\frac{1}{2}[{11}^2+3^2+\sqrt{{({11}^2+3^2)}^2-2{(7^2-{11}^2)}^2-2{(7^2-3^2)}^2}]=65+7\sqrt{17}$$$$\Rightarrow L=\sqrt{65+7\sqrt{17}}$$$$\cos \alpha =\frac{L^2+7^2-{11}^2}{2\times 7\times L}=\frac{\sqrt{17}-1}{2\sqrt{65+7\sqrt{17}}}$$$$\sin \alpha =\frac{L^2+7^2-3^2}{2\times 7\times L}=\frac{15+\sqrt{17}}{2\sqrt{65+7\sqrt{17}}}$$$$A_{orange}=\frac{L^2}{2}-\frac{L\times 7}{2}(\sin \alpha +\cos \alpha )$$$$=\frac{65+7\sqrt{17}}{2}-\frac{7}{2}\times (\frac{\sqrt{17}-1}{2}+\frac{15+\sqrt{17}}{2})$$$$=8$$

Commented by I want to learn more last updated on 16/Aug/20

$$\mathrm{Wow},\mathrm{i}\mathrm{really}\mathrm{appreciate}\mathrm{sir}$$

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