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Question Number 105883 by bemath last updated on 01/Aug/20

Answered by mr W last updated on 01/Aug/20

$$AG=\sqrt{3^2+6^2-2\times 3\times 6\times \cos 120°}=3\sqrt{7}$$$$\frac{\sin \mathrm{\angle }AGF}{3}=\frac{\sin 120°}{3\sqrt{7}}$$$$\sin \mathrm{\angle }AGF=\frac{\sqrt{21}}{14}$$$$\cos \mathrm{\angle }AGF=\frac{5\sqrt{7}}{14}$$$$HG\times \frac{\sqrt{21}}{14}=(3-HG\times \frac{5\sqrt{7}}{14})\tan 60°$$$$\Rightarrow HG=\sqrt{7}$$$$x=FH=3\sqrt{7}-\sqrt{7}=2\sqrt{7}$$

Commented by bemath last updated on 01/Aug/20

$$thankyousir$$

Answered by mr W last updated on 01/Aug/20

Commented by mr W last updated on 01/Aug/20

$$FL=\frac{3\sqrt{3}}{2}$$$$LG=3+3+\frac{3}{2}=\frac{15}{2}$$$$FG=\sqrt{{\left(\frac{3\sqrt{3}}{2}\right)}^2+{\left(\frac{15}{2}\right)}^2}=3\sqrt{7}$$$$\frac{HG}{FG}=\frac{BG}{KG}=\frac{1}{3}$$$$\Rightarrow HG=\frac{1}{3}FG$$$$x=FH=\frac{2}{3}FG=2\sqrt{7}$$

Answered by Coronavirus last updated on 01/Aug/20

$$\mathrm{en}\mathrm{appliquant}\mathrm{le}\mathrm{theoreme}{\mathrm{d}}^{\prime }\mathrm{Alkashi}\mathrm{au}\mathrm{triangle}\mathrm{A}GF$$$$alors{FG}^2={\mathrm{FA}}^2+\mathrm{A}{G}^2+2F\overrightarrow{A}.A\overrightarrow{G}$$$$={\left(3\right)}^2+{\left(6\right)}^2+2\left(3\right)\left(6\right)\cos (F\overrightarrow{A};A\overrightarrow{G})car\{\begin{array}{l}AG=AB+BG\\ BG=FG=ED=3\end{array}$$$$\mathrm{or}(\mathrm{F}\overrightarrow{A};A\overrightarrow{G})=\frac{\pi }{3}\Rightarrow \cos (\mathrm{F}\overrightarrow{A};A\overrightarrow{G})=\frac{1}{2}$$$$aussi{FG}^2=9+36+18=63$$$$doncFG=\sqrt{63}$$

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