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Question Number 205588 by cortano12 last updated on 25/Mar/24

Answered by A5T last updated on 25/Mar/24

Commented by A5T last updated on 25/Mar/24

$$BC=r\sqrt{2};{BD}^2=r^2+r^2-2r^{2}cos60\Rightarrow BD=r$$$$6^2+{AB}^2-6\sqrt{3}AB=r^2...\left(i\right)\Rightarrow 2r^2=72+2{AB}^2-12\sqrt{3}AB$$$$2r^2={BC}^2=96+{AB}^2-8\sqrt{3}AB=72+2{AB}^2-12\sqrt{3}AB$$$$\Rightarrow {AB}^2+4\sqrt{3}AB-24=0\Rightarrow AB=2(\sqrt{3}+3)\approx 9.464$$

Answered by mr W last updated on 25/Mar/24

Commented by mr W last updated on 25/Mar/24

$${CD}^2=6^2+{\left(4\sqrt{6}\right)}^2-2\times 6\times 4\sqrt{6}\times \cos 75°$$$$\Rightarrow CD=2\sqrt{15+6\sqrt{3}}$$$$\frac{4\sqrt{6}}{\sin \alpha }=\frac{2\sqrt{15+6\sqrt{3}}}{\sin 75°}=2R$$$$\sin \alpha =\frac{4\sqrt{6}\times \sin 75°}{2\sqrt{15+6\sqrt{3}}}=\frac{1+\sqrt{3}}{\sqrt{5+2\sqrt{3}}}$$$$\Rightarrow \cos \alpha =\frac{1}{\sqrt{5+2\sqrt{3}}}$$$$\frac{AB}{\sin (\alpha +45°)}=2R$$$$\Rightarrow AB=\frac{2\sqrt{15+6\sqrt{3}}}{\sin 75°}\times \frac{\sin \alpha +\cos \alpha }{\sqrt{2}}$$$$=2(3+\sqrt{3})✓$$

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