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(D) Let the magnitudes of the vectors be $|\overrightarrow{A}| = |\overrightarrow{B}| = x$ and $|\overrightarrow{C}| = x\sqrt{2}$.Given $\overrightarr

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$90^\circ, 135^\circ, 135^\circ$

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(D) Let the magnitudes of the vectors be $|\overrightarrow{A}| = |\overrightarrow{B}| = x$ and $|\overrightarrow{C}| = x\sqrt{2}$.Given $\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = 0$,which implies $\overrightarrow{A} + \overrightarrow{B} = -\overrightarrow{C}$.Squaring both sides: $|\overrightarrow{A}|^2 + |\overrightarrow{B}|^2 + 2|\overrightarrow{A}||\overrightarrow{B}|\cos(	heta_{AB}) = |\overrightarrow{C}|^2$.Substituting the values: $x^2 + x^2 + 2x^2\cos(	heta_{AB}) = (x\sqrt{2})^2 = 2x^2$.$2x^2 + 2x^2\cos(	heta_{AB}) = 2x^2$,which gives $\cos(	heta_{AB}) = 0$,so $	heta_{AB} = 90^\circ$.Since the vectors form a closed triangle,the exterior angles between the vectors are determined by the geometry of the triangle. The angles between the vectors are $90^\circ, 135^\circ, 135^\circ$.

Among three vectors,two vectors are equal in magnitude,and the magnitude of the third vector is $\sqrt{2}$ times the magnitude of the other two vectors. If $\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = 0$,find the angles between the vectors.

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