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Question Number 69092 by Mr. K last updated on 19/Sep/19

Answered by mr W last updated on 21/Sep/19

Commented by mr W last updated on 21/Sep/19

$$AX=\frac{a}{\cos \theta }$$$$XY=\frac{a-a\tan \theta }{\cos \theta }$$$$AX-XY=\frac{a\tan \theta }{\cos \theta }=\frac{a\sin \theta }{{\cos }^2\theta }$$$$\sqrt{{AX}^2+{(AX-XY)}^2}=\sqrt{\frac{a^2}{{\cos }^2\theta }+\frac{a^2{\sin }^2\theta }{{\cos }^4\theta }}$$$$=\sqrt{\frac{a^2({\cos }^2\theta +{\sin }^2\theta )}{{\cos }^4\theta }}=\frac{a}{{\cos }^2\theta }$$$$\frac{{AX}^2}{\sqrt{{AX}^2+{(AX-XY)}^2}}=\frac{{\left(\frac{a}{\cos \theta }\right)}^2}{\frac{a}{{\cos }^2\theta }}=a$$$$\Rightarrow proved!$$

Commented by TawaTawa last updated on 21/Sep/19

$$\mathrm{Wow},\mathrm{nice}\mathrm{sir}$$

Commented by Mr. K last updated on 21/Sep/19

$$Isthereanotherwaytoprove$$$$withoutusingtrigonometry?$$

Commented by mr W last updated on 22/Sep/19

$$ithinkso.forexampleyoucanset$$$$AB=a,BX=b.$$

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