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Question Number 208913 by efronzo1 last updated on 27/Jun/24

Answered by A5T last updated on 27/Jun/24

$$\frac{sin\beta }{6}=\frac{1}{AC};\frac{sin(60-\beta )}{3}=\frac{1}{AC}$$$$\Rightarrow \frac{sin(60-\beta )}{sin\beta }=\frac{1}{2}\Rightarrow \beta \approx 40.8934$$$$\Rightarrow sin\beta \approx 0.654654;wecanthenfindAC=\frac{6}{sin\beta }$$$$AC\approx 9.165151$$

Answered by mr W last updated on 27/Jun/24

$$\mathrm{\angle }BCD=90°-\alpha +90°-\beta =120°$$$${BD}^2=6^2+3^2-2\times 3\times 6\cos 120°=63$$$$AC=2R=\frac{BD}{\sin \mathrm{\angle }BCD}=\frac{\sqrt{63}}{\sin 120°}$$$$=2\sqrt{21}\approx 9.165$$

Answered by mr W last updated on 27/Jun/24

Commented by mr W last updated on 27/Jun/24

$$CE=2\times 6=12$$$$DE=12+3=15$$$$AD=\frac{15}{\sqrt{3}}=5\sqrt{3}$$$$AC=\sqrt{{\left(5\sqrt{3}\right)}^2+3^2}=2\sqrt{21}\approx 9.165$$

Commented by Tawa11 last updated on 27/Jun/24

$$\mathrm{weldone}\mathrm{sirs}$$

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