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Question Number 161811 by mnjuly1970 last updated on 22/Dec/21

Answered by mnjuly1970 last updated on 22/Dec/21

$$\mathrm{D}E=\sqrt{68}$$$$68=64+100-160cos(\mathrm{E}\stackrel{\wedge }{\mathrm{A}D}=\alpha )$$$$cos\left(\alpha \right)=\frac{96}{160}=\frac{3}{5}$$$$cos\left(\frac{\alpha }{2}\right)=\frac{8}{x}\Rightarrow \frac{1+\frac{3}{5}}{2}=\frac{64}{x^2}$$$$\frac{8}{10}=\frac{64}{x^2}\Rightarrow x^2=80\Rightarrow x=4\sqrt{5}$$$$$$$$$$

Answered by Rasheed.Sindhi last updated on 22/Dec/21

$$\mathrm{AB}=8$$$$\mathrm{AE}=\sqrt{8^2+6^2}=10$$$$\mathrm{\angle }\mathrm{BAE}=\theta$$$$\cos \theta =\frac{\mathrm{AB}}{\mathrm{AE}}=\frac{8}{10}=\frac{4}{5}$$$$\theta ={\cos }^{-1}\left(\frac{4}{5}\right)$$$$\mathrm{\angle }\mathrm{FAD}=\frac{90-\theta }{2}$$$$\mathrm{AD}=8$$$$\cos \mathrm{\angle }\mathrm{FAD}=\cos \frac{90-\theta }{2}=\frac{\mathrm{AD}}{\mathrm{AF}}=\frac{8}{\mathrm{x}}$$$$\mathrm{x}=\frac{8}{\cos \frac{90-\theta }{2}}=\frac{8}{\cos \left(\frac{90-{\cos }^{-1}\left(\frac{4}{5}\right)}{2}\right)}$$$$\mathrm{x}=\frac{8}{\cos \left(\frac{90-36.870}{2}\right)}\approx 8.94$$

Answered by mr W last updated on 22/Dec/21

$$\cos 2\theta =\frac{6}{10}=\frac{3}{5}$$$$\cos \theta =\sqrt{\frac{1+\cos 2\theta }{2}}=\sqrt{\frac{1+\frac{3}{5}}{2}}=\frac{2}{\sqrt{5}}$$$$\Rightarrow x=\frac{8}{\cos \theta }=\frac{8}{\frac{2}{\sqrt{5}}}=4\sqrt{5}$$

Commented by mnjuly1970 last updated on 23/Dec/21

$$veryexcellentsir\mathrm{W}$$

Commented by mr W last updated on 23/Dec/21

Commented by Tawa11 last updated on 22/Dec/21

$$\mathrm{great}\mathrm{sir}$$

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