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Question Number 56986 by mr W last updated on 27/Mar/19

Commented by mr W last updated on 27/Mar/19

$$Findx=?$$

Answered by MJS last updated on 28/Mar/19

Commented by mr W last updated on 28/Mar/19

$$correct,thankyousir!$$

Answered by mr W last updated on 28/Mar/19

Commented by mr W last updated on 28/Mar/19

$$alternativeway:$$$$AB=2\times 5=10$$$$AC=\sqrt{{10}^2-6^2}=8$$$$\cos \left(2\alpha \right)=\frac{AC}{AB}=\frac{8}{10}=\frac{4}{5}$$$$2{\cos }^2\alpha -1=\frac{4}{5}$$$$\Rightarrow \cos \alpha =\frac{3}{\sqrt{10}}$$$$AD=AC\cos \alpha =\frac{24}{\sqrt{10}}$$$$FD=DC=AC\sin \alpha =8\sqrt{1-\frac{9}{10}}=\frac{8}{\sqrt{10}}$$$$\frac{ED}{AD}=\frac{BC}{AC}=\frac{3}{4}$$$$\Rightarrow ED=\frac{3}{4}\times \frac{24}{\sqrt{10}}=\frac{18}{\sqrt{10}}$$$$x=ED-FD=\frac{18}{\sqrt{10}}-\frac{8}{\sqrt{10}}=\frac{10}{\sqrt{10}}=\sqrt{10}$$

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