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Question Number 144268 by aliibrahim1 last updated on 23/Jun/21

Answered by imjagoll last updated on 24/Jun/21

$$\mathrm{consider}\measuredangle \mathrm{CDA}=90°-\alpha \mathrm{and}$$$$\measuredangle \mathrm{CAD}=90°-\alpha \mathrm{so}\mathrm{x}=\mathrm{CD}=3\mathrm{cm}$$

Commented by som(math1967) last updated on 24/Jun/21

$$Is\measuredangle BAC=90°?$$

Answered by mr W last updated on 24/Jun/21

$$4^2=5^2+x^2-10x\cos 2\alpha$$$$\Rightarrow \cos 2\alpha =\frac{x^2+9}{10x}$$$${AD}^2=3^2+x^2-6x\cos 2\alpha$$$${AD}^2=3^2+x^2-6x\times \frac{x^2+9}{10x}$$$$\Rightarrow {AD}^2=\frac{2x^2+18}{5}$$$$2^2={AD}^2+4^2-8AD\cos \alpha$$$$\frac{2x^2+18}{5}+12=8AD\cos \alpha$$$$\frac{x^2+39}{5}=4AD\cos \alpha$$$${\left(\frac{x^2+39}{5}\right)}^2=8\times \frac{2x^2+18}{5}\times (\cos 2\alpha +1)$$$${(x^2+39)}^{2}x=8(x^2+9)(x^2+10x+9)$$$$\Rightarrow x=1,3,9$$$$onlyx=3isvalid.$$

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