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(C) Let the two sides of the triangle be $\vec{u} = \sqrt{3}(\vec{a} 	imes \vec{b})$ and $\vec{v} = \vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$.Using t

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$\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}$

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(C) Let the two sides of the triangle be $\vec{u} = \sqrt{3}(\vec{a} 	imes \vec{b})$ and $\vec{v} = \vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$.Using the vector triple product formula,$\vec{a} 	imes (\vec{b} 	imes \vec{a}) = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$.Since $\vec{a}$ is a unit vector,$\vec{a} \cdot \vec{a} = 1$,so $\vec{v} = \vec{a} 	imes (\vec{b} 	imes \vec{a})$.Note that $\vec{u}$ is perpendicular to $\vec{a}$ and $\vec{b}$,and $\vec{v}$ is perpendicular to $\vec{a}$.Specifically,$\vec{u} = \sqrt{3}(\vec{a} 	imes \vec{b})$ is perpendicular to $\vec{v} = \vec{a} 	imes (\vec{b} 	imes \vec{a})$ because $\vec{a} 	imes \vec{b}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$,while $\vec{a} 	imes (\vec{b} 	imes \vec{a})$ lies in the plane of $\vec{a}$ and $\vec{b}$.Thus,the angle between these two sides is $90^{\circ}$ or $\frac{\pi}{2}$.Let $	heta$ be the angle between the third side and side $\vec{v}$. Then $	an 	heta = \frac{|\vec{u}|}{|\vec{v}|}$.$|\vec{u}| = \sqrt{3} |\vec{a} 	imes \vec{b}| = \sqrt{3} |\vec{a}| |\vec{b}| \sin \phi = \sqrt{3} |\vec{b}| \sin \phi$,where $\phi$ is the angle between $\vec{a}$ and $\vec{b}$.$|\vec{v}| = |\vec{a} 	imes (\vec{b} 	imes \vec{a})| = |\vec{a}| |\vec{b} 	imes \vec{a}| \sin(90^{\circ}) = |\vec{b}| \sin \phi$.Therefore,$	an 	heta = \frac{\sqrt{3} |\vec{b}| \sin \phi}{|\vec{b}| \sin \phi} = \sqrt{3}$.This gives $	heta = \frac{\pi}{3}$.The third angle is $\pi - (\frac{\pi}{2} + \frac{\pi}{3}) = \frac{\pi}{6}$.Thus,the angles are $\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}$.

Let $\vec{a}$ be a unit vector and $\vec{b}$ be a nonzero vector not parallel to $\vec{a}$. The angles of the triangle,two of whose sides are represented by $\sqrt{3}(\vec{a} \times \vec{b})$ and $\vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$,are

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