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(A) Given waves are $y_1 = 5 \sin (\omega t - kx)$ and $y_2 = -5 \cos (\omega t - kx - 150^{\circ})$.We know that $-\cos(	heta) = \sin(	heta - 90^{\

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$5$

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(A) Given waves are $y_1 = 5 \sin (\omega t - kx)$ and $y_2 = -5 \cos (\omega t - kx - 150^{\circ})$.We know that $-\cos(	heta) = \sin(	heta - 90^{\circ})$.So,$y_2 = 5 \sin (\omega t - kx - 150^{\circ} - 90^{\circ}) = 5 \sin (\omega t - kx - 240^{\circ})$.Alternatively,$y_2 = 5 \sin (\omega t - kx - 240^{\circ} + 360^{\circ}) = 5 \sin (\omega t - kx + 120^{\circ})$.The resultant amplitude $A$ of two waves $A_1 \sin(\phi_1)$ and $A_2 \sin(\phi_2)$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\Delta\phi)}$.Here $A_1 = 5$,$A_2 = 5$,and the phase difference $\Delta\phi = 120^{\circ} - 0^{\circ} = 120^{\circ}$.$A = \sqrt{5^2 + 5^2 + 2(5)(5) \cos(120^{\circ})}$.Since $\cos(120^{\circ}) = -0.5$,we have $A = \sqrt{25 + 25 + 50(-0.5)} = \sqrt{50 - 25} = \sqrt{25} = 5$.

The resultant amplitude due to the superposition of two waves $y_1 = 5 \sin (\omega t - kx)$ and $y_2 = -5 \cos (\omega t - kx - 150^{\circ})$ is:

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