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(A) Given: Magnitude of vectors $A = B = 5$ units,angle $	heta = 60^\circ$.$(a)$ Magnitude of the sum of vectors: $R_s = \sqrt{A^2 + B^2 + 2AB \cos \

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$5\sqrt{3}$ units and $5$ units

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(A) Given: Magnitude of vectors $A = B = 5$ units,angle $	heta = 60^\circ$.$(a)$ Magnitude of the sum of vectors: $R_s = \sqrt{A^2 + B^2 + 2AB \cos 	heta} = \sqrt{5^2 + 5^2 + 2(5)(5) \cos 60^\circ} = \sqrt{25 + 25 + 50(0.5)} = \sqrt{75} = 5\sqrt{3}$ units.$(b)$ Magnitude of the difference of vectors: $R_d = \sqrt{A^2 + B^2 - 2AB \cos 	heta} = \sqrt{5^2 + 5^2 - 2(5)(5) \cos 60^\circ} = \sqrt{25 + 25 - 50(0.5)} = \sqrt{25} = 5$ units.

Two vectors of equal magnitude $5$ units have an angle $60^\circ$ between them. Find the magnitude of $(a)$ the sum of the vectors and $(b)$ the difference of the vectors.

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