Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the arc AC such that $$\frac{\text{lenght of arc AB}}{\text{lenght of arc BC}}=\frac{1}{5}$$,and $$\overrightarrow{OC}=\alpha\overrightarrow{OA}+\beta\overrightarrow{OB}$$, then $$\alpha +\sqrt{2}(\sqrt{3}-1)\beta$$ is equal to

Arc AC subtends $$90°$$ at center O. B divides arc AC in ratio 1:5, so arc AB subtends $$\frac{1}{6} \times 90° = 15°$$.

With A at angle $$0°$$, B at $$15°$$, C at $$90°$$ (unit radius):

$$\vec{OA} = (1, 0)$$, $$\vec{OB} = (\cos 15°, \sin 15°)$$, $$\vec{OC} = (0, 1)$$.

From $$\vec{OC} = \alpha\vec{OA} + \beta\vec{OB}$$:

$$0 = \alpha + \beta\cos 15°$$ and $$1 = \beta\sin 15°$$.

$$\beta = \frac{1}{\sin 15°} = \frac{4}{\sqrt{6}-\sqrt{2}} = \sqrt{6}+\sqrt{2}$$

$$\alpha = -\cot 15° = -(2+\sqrt{3})$$

$$(\sqrt{3}-1)(\sqrt{6}+\sqrt{2}) = 3\sqrt{2}-\sqrt{2} = 2\sqrt{2}$$

$$\sqrt{2}(\sqrt{3}-1)\beta = \sqrt{2} \times 2\sqrt{2} = 4$$

$$\alpha + \sqrt{2}(\sqrt{3}-1)\beta = -(2+\sqrt{3}) + 4 = 2 - \sqrt{3}$$