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Question Number 123027 by Lekhraj last updated on 21/Nov/20

Commented by kolos last updated on 24/Nov/20

$$dude,howdidyoupostafigure?$$

Answered by TANMAY PANACEA last updated on 21/Nov/20

$$AC=y$$$$BC=x$$$${\left(AC\right)}^2+{\left(BC\right)}^2={\left(AB\right)}^2$$$$angle<ABC=\theta AB=hypotsneous$$$$AC=perpendicularBC=base$$$$tan\theta =\frac{AC}{BC}tan(90-\theta )=\frac{BC}{AC}$$$$sin\theta =\frac{CD}{BC}sin(90-\theta )=\frac{CD}{AC}$$$${sin}^2\theta +{cos}^2\theta =1$$$$\frac{{CD}^2}{{BC}^2}+\frac{{CD}^2}{{AC}^2}=1$$$$\frac{1}{{BC}^2}+\frac{1}{{AC}^2}=\frac{1}{{CD}^2}$$$$$$

Answered by Olaf last updated on 21/Nov/20

$$\mathrm{Area}=\frac{1}{2}AC\times BC=\frac{1}{2}AB\times CD$$$$\Rightarrow {CD}^2=\frac{{AC}^2\times {BC}^2}{{AB}^2}\left(1\right)$$$$\mathrm{and}\mathrm{Pythagore}:{AC}^2+{BC}^2={AB}^2$$$$\left(1\right):\frac{1}{{CD}^2}=\frac{{AC}^2+{BC}^2}{{AC}^2\times {BC}^2}=\frac{1}{{BC}^2}+\frac{1}{{AC}^2}$$$$\left(\mathrm{Descartes}\mathrm{formula}\right)$$

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