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Question Number 76087 by aliesam last updated on 23/Dec/19

Answered by mr W last updated on 23/Dec/19

Commented by mr W last updated on 23/Dec/19

$$AB=\sqrt{2^2+6^2}=2\sqrt{10}$$$$AC=\sqrt{10}$$$$AO=R=\sqrt{50}$$$$\cos \mathrm{\angle }OAC=\frac{\sqrt{10}}{\sqrt{50}}=\frac{1}{\sqrt{5}}$$$$\cos \mathrm{\angle }DAC=\frac{6}{2\sqrt{10}}=\frac{3}{\sqrt{10}}$$$$\cos \alpha =\cos (\mathrm{\angle }OAC-\mathrm{\angle }DAC)$$$$=\frac{1}{\sqrt{5}}\times \frac{3}{\sqrt{10}}+\frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{10}}=\frac{\sqrt{2}}{2}$$$$x^2={OD}^2={\left(\sqrt{50}\right)}^2+6^2-2\sqrt{50}\times 6\times \cos \alpha$$$$=50+36-2\sqrt{50}\times 6\times \frac{\sqrt{2}}{2}=26$$$$\Rightarrow x=\sqrt{26}$$

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