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Question Number 81122 by mr W last updated on 09/Feb/20

Commented by mr W last updated on 09/Feb/20

$$blackdotisthecenterofcircle.$$$$bothanglesmarkedwith‘‘{\mathrm{o}}^{″}areequal.$$$$twolengthesaregiven:5and6.$$$$findlengthx=?$$

Commented by ajfour last updated on 09/Feb/20

Commented by ajfour last updated on 09/Feb/20

$$\sin 2\alpha =\frac{3}{5},\cos 2\alpha =\frac{4}{5}$$$$\tan \alpha =\sqrt{\frac{1/5}{9/5}}=\frac{1}{3}$$$$sayradius=5=r$$$$q=\left(2r\cos 2\alpha \right)\cos \alpha \tan \alpha$$$$q+x=\left(2r\cos 2\alpha \right)\cos \alpha \tan 2\alpha$$$$x=q+x-q$$$$=\left(10\times \frac{4}{5}\times \frac{3}{\sqrt{10}}\right)(\frac{3}{4}-\frac{1}{3})$$$$x=\frac{24}{\sqrt{10}}\times \frac{5}{12}=\sqrt{10}.$$

Commented by mr W last updated on 09/Feb/20

$$correctanswer!thankssir!$$

Answered by mr W last updated on 09/Feb/20

Commented by mr W last updated on 09/Feb/20

$$DB=2\times 5=10$$$$BC=\sqrt{{10}^2-6^2}=8$$$$\cos 2\alpha =\frac{8}{10}=\frac{4}{5}=2{\cos }^2\alpha -1$$$$\Rightarrow \cos \alpha =\frac{3}{\sqrt{10}}\Rightarrow \tan \alpha =\frac{1}{3}$$$$BE=BC\times \cos \alpha =8\times \frac{3}{\sqrt{10}}=\frac{24}{\sqrt{10}}$$$$AE=BE\times \tan 2\alpha$$$$FE=BE\times \tan \alpha$$$$x=AE-FE=BE\times (\tan 2\alpha -\tan \alpha )$$$$=\frac{24}{\sqrt{10}}(\frac{3}{4}-\frac{1}{3})=\sqrt{10}$$

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