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Question Number 12846 by Joel577 last updated on 04/May/17

Commented by Joel577 last updated on 04/May/17

$$HowtofindCE?$$

Answered by mrW1 last updated on 04/May/17

$$FB=12/2=6$$$$\tan \alpha =6/12=1/2$$$$EC=\frac{DC}{\cos \beta }=\frac{12}{\cos (90-2\alpha )}=\frac{12}{\sin \left(2\alpha \right)}$$$$\sin \left(2\alpha \right)=\frac{2\tan \alpha }{1+{\tan }^2\alpha }=\frac{2\times \frac{1}{2}}{1+{\left(\frac{1}{2}\right)}^2}=\frac{4}{5}$$$$\Rightarrow EC=\frac{12}{\frac{4}{5}}=15$$$$$$

Commented by mrW1 last updated on 04/May/17

$$yes,CG=12.GE=EA=3.$$

Commented by mrW1 last updated on 04/May/17

Commented by Joel577 last updated on 04/May/17

$$thankyouverymuch$$

Commented by Joel577 last updated on 04/May/17

$$butwithyourfigure,isCG=12?$$$$becauseIthought\mathrm{\Delta }CFG\cong \mathrm{\Delta }BCF$$

Commented by mrW1 last updated on 04/May/17

$$youcanalsosolvelikethis:$$$$GE=EA=x$$$$\frac{GE}{FG}=\frac{FG}{CG}$$$$\frac{x}{6}=\frac{6}{12}$$$$x=3$$$$CE=12+3=15$$

Commented by chux last updated on 04/May/17

$$\mathrm{wow}.....\mathrm{i}\mathrm{love}\mathrm{this}$$$$$$

Commented by Joel577 last updated on 04/May/17

$$thankyouverymuch$$

Commented by A Haq Soomro last updated on 06/May/17

$$\eta \stackrel{\bullet }{ı}\subset \in !$$

Answered by ajfour last updated on 04/May/17

Commented by ajfour last updated on 04/May/17

$$BC=CG=12(tangentlength$$$$\left(fromsamepointtosamecircle\right)$$$$GE=AE=x$$$$DE=12-x$$$$in△CDE,{CD}^2+{DE}^2={CE}^2$$$${12}^2+{(12-x)}^2={(12+x)}^2$$$$144=4\left(12\right)x\Rightarrow x=3$$$$CE=12+x=12+3=15.$$

Commented by chux last updated on 04/May/17

$$\mathrm{i}\mathrm{love}\mathrm{this}....\mathrm{thanks}\mathrm{boss}$$

Commented by A Haq Soomro last updated on 06/May/17

$$Excellent!$$

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