Let the arc AC of a circle subtend a right an | vedclass

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(A) Let $\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and $\overrightarrow{OC} = \vec{c}$. Since $A, B, C$ lie on a circle with cent

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$2-\sqrt{3}$

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(A) Let $\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and $\overrightarrow{OC} = \vec{c}$. Since $A, B, C$ lie on a circle with centre $O$,$|\vec{a}| = |\vec{b}| = |\vec{c}| = R$. Given that arc $AC$ subtends $90^{\circ}$ at the centre,the angle between $\vec{a}$ and $\vec{c}$ is $90^{\circ}$. Since the ratio of arc lengths $AB:BC = 1:5$,the angle $\angle AOB = \frac{1}{1+5} 	imes 90^{\circ} = 15^{\circ}$ and $\angle BOC = \frac{5}{6} 	imes 90^{\circ} = 75^{\circ}$. We have $\vec{c} = \alpha \vec{a} + \beta \vec{b}$. Taking the dot product with $\vec{a}$: $\vec{a} \cdot \vec{c} = \alpha |\vec{a}|^2 + \beta \vec{a} \cdot \vec{b} \Rightarrow R^2 \cos 90^{\circ} = \alpha R^2 + \beta R^2 \cos 15^{\circ} \Rightarrow 0 = \alpha + \beta \cos 15^{\circ} \Rightarrow \alpha = -\beta \cos 15^{\circ} \dots (1)$. Taking the dot product with $\vec{b}$: $\vec{b} \cdot \vec{c} = \alpha \vec{a} \cdot \vec{b} + \beta |\vec{b}|^2 \Rightarrow R^2 \cos 75^{\circ} = \alpha R^2 \cos 15^{\circ} + \beta R^2 \Rightarrow \cos 75^{\circ} = \alpha \cos 15^{\circ} + \beta \dots (2)$. Substituting $(1)$ into $(2)$: $\cos 75^{\circ} = -\beta \cos^2 15^{\circ} + \beta = \beta(1 - \cos^2 15^{\circ}) = \beta \sin^2 15^{\circ}$. Thus,$\beta = \frac{\cos 75^{\circ}}{\sin^2 15^{\circ}} = \frac{\sin 15^{\circ}}{\sin^2 15^{\circ}} = \frac{1}{\sin 15^{\circ}} = \frac{1}{\frac{\sqrt{6}-\sqrt{2}}{4}} = \frac{4}{\sqrt{6}-\sqrt{2}} = \sqrt{6}+\sqrt{2}$. From $(1)$,$\alpha = -\beta \cos 15^{\circ} = -(\sqrt{6}+\sqrt{2}) \left(\frac{\sqrt{6}+\sqrt{2}}{4}ight) = -\frac{6+2+2\sqrt{12}}{4} = -\frac{8+4\sqrt{3}}{4} = -(2+\sqrt{3})$. We need to evaluate $\alpha + \sqrt{2}(\sqrt{3}-1) \beta = -(2+\sqrt{3}) + \sqrt{2}(\sqrt{3}-1)(\sqrt{6}+\sqrt{2}) = -(2+\sqrt{3}) + \sqrt{2}(\sqrt{18}+\sqrt{6}-\sqrt{6}-\sqrt{2}) = -(2+\sqrt{3}) + \sqrt{2}(3\sqrt{2}-\sqrt{2}) = -(2+\sqrt{3}) + \sqrt{2}(2\sqrt{2}) = -(2+\sqrt{3}) + 4 = 2-\sqrt{3}$.

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