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Question Number 147259 by mathdanisur last updated on 19/Jul/21

Answered by Rasheed.Sindhi last updated on 19/Jul/21

$$△AED:$$$$\because ADisdiameter$$$$\therefore \mathrm{\angle }AED=90$$$$tan\mathrm{\angle }A=\frac{1}{3}\Rightarrow \mathrm{\angle }A=19.47$$$$AC=\sqrt{{AE}^2+{EC}^2}=\sqrt{3^2+1^2}=\sqrt{10}$$$$△ABE:$$$$AB=AC=\sqrt{10}$$$$\mathrm{\angle }BAE=90+19.47=109.47$$$${EB}^2={AB}^2+{AE}^2-2AB.AE.cos\mathrm{\angle }BAE$$$${EB}^2={\left(\sqrt{10}\right)}^2+{\left(3\right)}^2-2\sqrt{10}.3.cos109.47$$$$=10+9-6\sqrt{10}(-\frac{1}{3})$$$$=19+2\sqrt{10}$$$$EB=\sqrt{19+2\sqrt{10}}\approx 5.03$$

Commented by mathdanisur last updated on 19/Jul/21

$$thankyouSer$$

Commented by otchereabdullai@gmail.com last updated on 19/Jul/21

$$\mathrm{nice}!$$

Answered by liberty last updated on 19/Jul/21

$$AD=\sqrt{3^2+1^2}=\sqrt{10}=AB$$$$let\mathrm{\angle }EAD=\alpha$$$$\cos \alpha =\frac{9+10-1}{2.3.\sqrt{10}}=\frac{3}{\sqrt{10}}then$$$$\sin \alpha =\frac{1}{\sqrt{10}}$$$$\mathrm{\angle }EAB=90°+\alpha$$$$\cos (90°+\alpha )=\frac{9+10-{EB}^2}{2.3.\sqrt{10}}$$$$\Rightarrow -\sin \alpha =\frac{19-{EB}^2}{6\sqrt{10}}$$$$\Rightarrow -\frac{1}{\sqrt{10}}\times 6\sqrt{10}=19-{EB}^2$$$$\Rightarrow EB=\sqrt{19+6}=5$$

Commented by mathdanisur last updated on 19/Jul/21

$$thankyouSer$$

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