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(B) Given: Magnitudes of vectors $P = x$ and $Q = x$. The angle between them is $	heta = 45^\circ$. The resultant magnitude is $R = \sqrt{2 + \sqrt{2

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$1$

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(B) Given: Magnitudes of vectors $P = x$ and $Q = x$. The angle between them is $	heta = 45^\circ$. The resultant magnitude is $R = \sqrt{2 + \sqrt{2}}$.Using the vector addition formula: $R = \sqrt{P^2 + Q^2 + 2PQ \cos 	heta}$.Substituting the values: $R = \sqrt{x^2 + x^2 + 2(x)(x) \cos 45^\circ}$.Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$,we have $R = \sqrt{2x^2 + 2x^2 \left(\frac{1}{\sqrt{2}}ight)}$.Simplifying the expression: $R = \sqrt{2x^2 + \sqrt{2}x^2} = \sqrt{x^2(2 + \sqrt{2})} = x\sqrt{2 + \sqrt{2}}$.Equating this to the given resultant: $x\sqrt{2 + \sqrt{2}} = \sqrt{2 + \sqrt{2}}$.Therefore,$x = 1$.

Two vectors having equal magnitudes of $x$ units acting at an angle of $45^\circ$ have a resultant of $\sqrt{2 + \sqrt{2}}$ units. The value of $x$ is

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