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Question Number 199089 by mr W last updated on 28/Oct/23

Answered by ajfour last updated on 28/Oct/23

Commented by ajfour last updated on 28/Oct/23

$${OB}^2={(2\sqrt{10}-2\cos \alpha )}^2+{(\sqrt{10}-2\sin \alpha )}^2$$$$as\tan \alpha =\frac{1}{3}$$$${OB}^2={(\frac{20}{\sqrt{10}}-\frac{6}{\sqrt{10}})}^2+{(\frac{10}{\sqrt{10}}-\frac{2}{\sqrt{10}})}^2$$$$=\frac{196+64}{10}=26$$$$OB=\sqrt{26}$$

Commented by mr W last updated on 28/Oct/23

$$veryniceapproach!$$$$thanksalot!$$

Answered by mr W last updated on 28/Oct/23

$$AC=\sqrt{6^2+2^2}=2\sqrt{10}$$$$\cos \mathrm{\angle }OAC=\frac{AC}{2\times OA}=\frac{\sqrt{10}}{\sqrt{50}}=\frac{1}{\sqrt{5}}$$$$\cos \mathrm{\angle }BAC=\frac{6}{2\sqrt{10}}=\frac{3}{\sqrt{10}}$$$$\cos \mathrm{\angle }OAB=\cos (\mathrm{\angle }OAC-\mathrm{\angle }BAC)$$$$=\frac{1}{\sqrt{5}}\times \frac{3}{\sqrt{10}}+\frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{10}}=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}}$$$${OB}^2=6^2+{\left(\sqrt{50}\right)}^2-2\times 6\times \sqrt{50}\times \frac{1}{\sqrt{2}}=26$$$$\Rightarrow OB=\sqrt{26}$$

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